The rational homotopy type of the complement of the graph of a map

Abstract

We prove that the rational homotopy type of the complement of the graph of a continuous map from a simply connected closed manifold to a 2-connected closed manifold of the same dimension depends only on the rational homotopy class of the map. We give a commutative differential graded algebra model (in the sense of Sullivan) of the complement of the graph and study its formality.

1 Introduction

In this paper we have considered only smooth manifolds. However using the results of Milnor [14], Kister [8] and Kirby–Siebenmann [7] the arguments given for smooth manifolds can be carried over to topological manifolds.

Let $f:M \rightarrow N$ be a continuous map of closed connected oriented manifolds of dimension $n$. We aim to study the rational homotopy of $M \times N \backslash \Gamma (f)$, where $\Gamma (f)=\{(x,f(x))|x\in M\}$ is the graph of $f$.

We refer the reader to [3] and [5] for standard definitions and terminologies of: Free commutative differential graded algebra (CDGA), (minimal) Sullivan CDGA, (minimal) Sullivan model for a CDGA and for a space X, homotopy between morphisms of CDGAs, formality of spaces.

Definition 1.1

([5], [4, 22], [12, 20]) A model for a continuous map $f:X\rightarrow Y$ between connected topological spaces is a morphism of CDGAs, $\psi _f:(A_Y,d_{A_Y})\rightarrow (A_X,d_{A_X})$, such that there exists a homotopy commutative diagram, where $(\Lambda W, d_W),(\Lambda V, d_V)$ are minimal models of $Y$ and $X$ respectively and the vertical arrows are quasi-isomorphisms:

$$\begin{aligned} \left. \begin{array}{lll} (A_Y,d_{A_Y})&{}\buildrel {\psi _f}\over \rightarrow &{} (A_X,d_{A_X})\\ \cong _q \;\uparrow \; \phi _W &{}&{} \; \cong _q \;\uparrow \; \phi _V \\ (\Lambda W, d_W) &{} \buildrel {\psi _{\Lambda f}}\over \rightarrow &{} (\Lambda V, d_V) \\ \; \cong _q \;\downarrow \; \rho _W &{}&{} \; \cong _q \;\downarrow \; \rho _V \\ (A_{PL}(Y), d_Y) &{} \buildrel {A_{PL}(f)}\over \rightarrow &{} (A_{PL}(X), d_X). \end{array}\right. \end{aligned}$$

A continuous map $f:X\rightarrow Y$ between formal spaces is said to be formal (also referred to as formalizable in the literature) if $f^*:(H^*(Y; \mathbb {Q}), 0) \rightarrow (H^*(X; \mathbb {Q}), 0)$ is a model for $f$.

Many authors have studied rational homotopy type of maps and their formality; see e.g. [4, 13, 16–18, 22].

These authors have studied spaces $C(f)$ closely associated with $f$, like the mapping cone of $f$, the cofibre of $f$, the homotopy fibre of $f$ and have recorded results related to problems of following types:

Problem 1.2

• 1. Find out if the rational homotopy of $C(f)$ is determined by the rational homotopy of $f$.

• 2. Given that $f$ is formal, determine if $C(f)$ is necessarily formal.

We study the following problems of a similar nature:

Problem 1.3

• 1. Find out if the rational homotopy of $M \times N \backslash \Gamma (f)$, where $\Gamma (f)$ is the graph of $f$, is determined by the rational homotopy of $f$.

• 2. Given that $f$ is formal, determine if $M \times N \backslash \Gamma (f)$ is necessarily formal.

The motivation of this paper comes from the work of Lambrechts and Stanley [9] on the rational homotopy type of configuration spaces of two points, and the observation that if $f$ is the identity map $1_M:M \rightarrow M$ then, the graph of $f$, $\Gamma (f) = \Delta$, the diagonal, and $M \times M \backslash \Gamma (f) = M \times M \backslash \Delta = F(M, 2)$, the configuration space of two points. We begin with the following immediate observations:

• 1. If $f: {\mathbb {R}}^n \rightarrow {\mathbb {R}}^n$ is any continuous map then ${\mathbb {R}}^n \times {\mathbb {R}}^n \backslash \Gamma (f)$ is homeomorphic to $F({\mathbb {R}}^n, 2)$ under the homeomorphism $(u, v) \mapsto (u, v-f(u)+u)$.

• 2. For a topological space $X$ if $f:X \rightarrow X$ is a homeomorphism then $X \times X \backslash \Gamma (f)$ is homeomorphic to $X \times X \backslash \Delta = F(X, 2)$ under the homeomorphism $(x, y) \mapsto (x, f^{-1}(y))$.

• 3. Suppose that $X = S^2\vee S^3$, $f:X\rightarrow X$ is the constant map $x\mapsto (0,0,1)$ and $g:X\rightarrow X$ is the constant map $x\mapsto (0,0,0,-1)$. Then $X \times X\setminus \Gamma (f)\simeq X \times (\mathbb {R}^2\vee S^3)$ whereas $X \times X\setminus \Gamma (g)\simeq X \times (S^2 \vee \mathbb {R}^3)$. Therefore they are not homotopy equivalent. Thus for an arbitrary space $X$ and a continuous map $f:X \rightarrow X$, $X \times X \backslash \Gamma (f)$ need not even be homotopy equivalent to $F(X, 2)$.

Lambrechts and Stanley [9, 10] proved that if $M$ is a simply-connected closed manifold such that $H^2(M;{\mathbb {Q}}) = 0$ then the rational homotopy type of $F(M,2)$ depends only on the rational homotopy type of $M$. They also proved in [9] that if $M$ is a 2-connected closed manifold, then $F(M,2)$ is formal if and only if $M$ is formal by constructing explicitly a CDGA model for $F(M,2)$ out of a differential Poincaré duality algebra model of $M$ and a model of the diagonal map.

We show that starting from a CDGA model of $f: M \rightarrow N$ one can construct a CDGA model of $(1_M \times f) \circ \Delta _M: M \rightarrow M \times N$, where $\Delta _M (x) = (x, x)$ for $x\in M$. This allows us by Corollary 1.5 of [10], to conclude that if we take $M, \;N$ to be simply connected and $H^2(N;\mathbb {Q}) =0$ then the rational homotopy type of $M \times N \setminus \Gamma (f)$ is determined by the rational homotopy class of $f$:

Theorem 1.4

Let $f:M \rightarrow N$ be a continuous map of closed connected oriented manifolds of dimension $n$ such that $H^1(M;\mathbb {Q}) = 0 = H^1(N;\mathbb {Q}) = H^2(N;\mathbb {Q}),$ then a CDGA-model of $M \times N\setminus \Gamma (f)$ can be explicitly determined out of any CDGA-model of $f$.

Corollary 1.5

If $f:M \rightarrow N$ is a continuous map of closed simply connected manifolds of dimension $n$ such that $H^2(N;\mathbb {Q}) =0,$ then the rational homotopy type of $M \times N \backslash \Gamma (f)$ depends only on the rational homotopy class of $f$.

We relativize results of [9] by defining a class $\Delta _{\psi }$ depending on a morphism $\psi$ of differential Poincaré duality algebras similar to the diagonal class $\Delta$ in [9]. We know from Theorem 1.1 of [11] that if $M, \; N$ are closed connected oriented manifolds with $H^1(M; {\mathbb {Q}}) = 0 = H^1(N; {\mathbb {Q}}),$ then they admit differential Poincaré duality algebra models. We also know from Proposition 1 of [2] that if $f:M \rightarrow N$ is a continuous map of simply connected manifolds of dimension $n \ge 7$ (it is a conjecture that this dimensional restriction can be removed) such that $H^2(f)$ is injective, then $f$ admits a Sullivan model $\psi _f:(A_N,d_{A_N})\rightarrow (A_M,d_{A_M})$, where $(A_M,d_{A_M}),\;(A_N,d_{A_N})$ are Poincaré duality algebras. Thus assuming the existence of such a model of $f$ we can construct a specific CDGA model of $M \times N\!\setminus \!\Gamma (f)$ from the model of $f$:

Theorem 1.6

Let $f:M \rightarrow N$ be a continuous map of closed connected oriented manifolds of dimension $n$ such that $H^1(M;\mathbb {Q}) = 0 = H^1(N;\mathbb {Q}) = H^2(N;\mathbb {Q}).$ If $\psi _f:(A_N,d_{A_N})\rightarrow (A_M,d_{A_M})$ is a model of $f,$ where $(A_M,d_{A_M})$, $(A_N,d_{A_N})$ are oriented differential Poincaré duality algebras, $\{a_i\}_{1\le i \le l}$ a homogeneous basis of $A_N$ and $\{a^*_i\}_{1\le i \le l}$ its Poincaré dual basis

$$\begin{aligned}\Delta = \sum _{i=1}^{l_N} (-1)^{\deg \; (a_i)} a_i \otimes a_i^* \in (A_N \otimes A_N)^n, \end{aligned}$$

is the diagonal class (defined in [9, p. 1030) and ${\Delta }_{\psi _f} := (\psi _f \otimes 1)(\Delta )$, called the class of ${\psi _f}$, then the ideal $({\Delta }_{\psi _f}) = {\Delta }_{\psi _f}.(A_M \otimes A_N)$ is a differential ideal of $A_M \otimes A_N$, and the quotient CDGA

$$\begin{aligned}\left( \frac{A_M \otimes A_N}{({\Delta }_{\psi _f})},\,\, \overline{d_{A_M} \otimes d_{A_N}}\right) \end{aligned}$$

is a CDGA model of $M \times N \backslash \Gamma (f)$.

This allows us to prove that if $f$ is formal then $M \times N\!\setminus \!\Gamma (f)$ is also formal.

Corollary 1.7

Let $f:M \rightarrow N$ be a formal map of closed connected oriented formal manifolds of dimension $n$ such that $H^1(M;\mathbb {Q}) = 0 = H^1(N;\mathbb {Q}) = H^2(N;\mathbb {Q}),$ then $M \times N\setminus \Gamma (f)$ is a formal space.

As the first major step in this paper we compute the cohomology algebra $H^*(M \times N \backslash \Gamma (f))$ using the Leray spectral sequence as described in §2 of Totaro [21]. We analyze at depth the cohomology class $\mu _f$, as described in Chapter 30 of Greenberg and Harper [6].

Let $\tilde{\mu } \in H^n(M \times N, M \times N\!\setminus \!\Gamma (f); \mathbb {Z})$ be the Thom class as in Lemma 2.1, and $\tilde{\mu '} = j^*(\tilde{\mu }) \in H^n(M \times N; \mathbb {Z})$, where $j: M \times N \rightarrow (M \times N, M \times N \setminus \Gamma (f))$ is the natural injection. Let us now take cohomology with rational coefficients (or coefficients in any field). Let $\{b_i\}_{1 \le i \le {l_N}}$ be a homogeneous basis of $H^*(N; {\mathbb {Q}})$, and let $\{b_i^*\}_{1 \le i \le {l_N}}$ be its Poincaré dual basis, that is $\langle b_i \cup b_j^*, [N]\rangle = \delta _{ij}$, where $[N]$ is the fundamental homology class of $N$. Let $\mu '_N := \sum _{i=1}^{l_N} (-1)^{\deg \; (b_i)} b_i \times b_i^* \in (H^*(N \times N; {\mathbb {Q}}))^n$ be the diagonal class of $N$, and let $\mu _f := (f \times id_N)^*(\mu '_N)$, called the cohomology class of the graph of $f$ (see p. 284 of [6]). Then there is a unique class $\Delta _f \in (H^*(M; {\mathbb {Q}})\otimes H^*(N; {\mathbb {Q}}))^n$ which maps to $\mu _f$ under the isomorphism of the Künneth theorem $(H^*(M; {\mathbb {Q}})\otimes H^*(N; {\mathbb {Q}}))^n \xrightarrow {\cong } (H^*(M \times N; {\mathbb {Q}}))^n$, we call $\Delta _f$ the class of $f$.

Theorem 1.8

Let $f:M \rightarrow N$ be a continuous map of closed oriented manifolds of dimension $n\ge 2$ and $\Gamma (f)$ be the graph of $f$. Then we have the following isomorphism of rings:

$$\begin{aligned} H^*(M \times N \backslash \Gamma (f); \mathbb {Z}) \cong \frac{H^*(M \times N; \mathbb {Z})}{(\tilde{\mu }')}, \end{aligned}$$

where $(\tilde{\mu }')$ denotes the ideal $H^*(M \times N; \mathbb {Z}) \cup \tilde{\mu }'$. For coefficients in $\mathbb {Q}$, and $\Delta _f =\sum ^{l_N}_{i=1}(-1)^{\deg \;(b_i)}f^*(b_i)\otimes b_i^{*}$, we have the following isomorphism of algebras:

$$\begin{aligned} H^*(M \times N\!\setminus \!\Gamma (f); {\mathbb {Q}}) \cong \frac{H^*(M \times N; {\mathbb {Q}})}{(\mu _f)} \cong \frac{H^*(M; {\mathbb {Q}})\otimes H^*(N; {\mathbb {Q}})}{(\Delta _f)}. \end{aligned}$$

The paper is arranged as follows. In §2 we determine the cohomology algebra of $M \times N \backslash \Gamma (f)$. In §3 we adapt for a continuous map $f:M\rightarrow N$ some results of [9] and [10] to construct a CDGA model of $M \times N \backslash \Gamma (f)$, and prove Theorem 1.4 and Corollary 1.5. In §4 we prove several results about differential Poincaré duality algebra associated to a morphism $\psi$ by adapting similar results of [9], proved in the absolute case. These results and some results of [10] are used for the proofs of Theorem 1.6 and Corollary 1.7. In the final §5 we give some examples and applications of our results.

2 The cohomology algebra of $M \times N \backslash \Gamma (f)$

In this section we prove Theorem 1.8.

We assume that $f:M \rightarrow N$ is a continuous map of closed connected oriented manifolds of dimension $n$ with graph $\Gamma (f)$. Since $\Gamma (f) \subset M \times N$ is an embedding, one can derive the following analogue of Corollary 11.2 of Milnor-Stasheff [15] (and also of Corollary (30.2) of Greenberg-Harper [6]) in a similar fashion. We follow the notations which preceded the statement of Theorem 1.8 in §1.

Lemma 2.1

There is a Thom class $\tilde{\mu } \in H^n(M \times N, M \times N \backslash \Gamma (f); {\mathbb {Z}})$ associated to the oriented normal bundle of the embedding $\Gamma (f) \subset M \times N$.

We next prove the following analogue of Corollary (30.3) of [6]:

Theorem 2.2

Let $[N] \in H_n(N; {\mathbb {Z}})$ be the fundamental homology class of $N$, $j^*:H^n(M \times N, M \times N \backslash \Gamma (f); {\mathbb {Z}}) \rightarrow H^n(M \times N; {\mathbb {Z}})$ be the homomorphism in cohomology induced by the map $j:M \times N \hookrightarrow (M \times N, M \times N \backslash \Gamma (f))$ and $\tilde{\mu }' = j^* (\tilde{\mu })$, then $\tilde{\mu }' / [N] = 1$.

Proof

For $x \in M$, consider the commutative diagram:

where $\tilde{i_x}:(N, N \backslash f(x)) \rightarrow (M \times N, M \times N \backslash \Gamma (f))$ is defined by $\tilde{i_x} (x') = (x, x'), \forall x' \in N$, $\tilde{i_x}: N \rightarrow M \times N$ its restriction to $(N, \emptyset )$ and $j_{f(x)}\!:\!N\rightarrow (N,N\!\setminus \!f(x))$ is the natural injection. If $s: N \rightarrow N^0$ is a section of the orientation sheaf $N^0$ over $N$, then

$$\begin{aligned}&1 = [s(f(x)), \tilde{i_x}^* (\tilde{\mu })] = [{j_{f(x)}}_*[N], \tilde{i_x}^* (\tilde{\mu })] = [[N], j_{f(x)}^* \tilde{i_x}^* (\tilde{\mu })] \\&\quad = [[N], (\tilde{i_x} \circ j_{f(x)})^* (\tilde{\mu })] = [[N], (j \circ \tilde{i_x} )^* (\tilde{\mu })] = [ [N], \tilde{i_x}^* \circ j^*(\tilde{\mu })] \\&\quad = [\tilde{i_x}_*[N],j^*(\tilde{\mu })] = [\tilde{i_x}_*[N], \tilde{\mu }']. \end{aligned}$$

Since $N \cong \{x\} \times N$, if $\bar{x}$ is the homology class of the zero cycle $x$, $\tilde{i_x}_*([N]) = \bar{x} \times [N]$. So from the above expression we get

$1 = [\tilde{i_x}_*[N], \tilde{\mu }'] = [\bar{x} \times [N], \tilde{\mu }'] = [\bar{x}, \tilde{\mu }' / [N]]$ for all $x \in M$, which proves the result. $\square$

We now prove the following analogue of Lemma 11.8 of [15] (and also of Lemma (30.5) of [6]):

Theorem 2.3

If $a \in H^*(N; {\mathbb {Z}})$, then $(f^*(a) \times 1) \cup \tilde{\mu }' = (1 \times a) \cup \tilde{\mu }'.$

Proof

Let $V_{\Gamma (f)}$ be a tubular neighborhood of the graph $\Gamma (f)$ in $M \times N$. So $\Gamma (f)$ is a deformation retract of $V_{\Gamma (f)}$. Let $i_{\Gamma (f)}:\Gamma (f) \rightarrow V_{\Gamma (f)}$ be the inclusion map and $r_{\Gamma (f)}:V_{\Gamma (f)}\rightarrow \Gamma (f)$ be the retraction map. Then $r_{\Gamma (f)} \circ i_{\Gamma (f)} = 1_{\Gamma (f)}$ and $i_{\Gamma (f)} \circ r_{\Gamma (f)} \simeq 1_{V_{\Gamma (f)}}$. Consider the projections $p_1:M \times N \rightarrow M$ and $p_2:M \times N \rightarrow N$. Since $f \circ p_1$ and $p_2$ coincide on $\Gamma (f)$, we have $(f \circ p_1)\mid _{V_{\Gamma (f)}} \circ \; i_{\Gamma (f)} \circ r_{\Gamma (f)}= p_2 \mid _{V_{\Gamma (f)}}\circ \;i_{\Gamma (f)} \circ r_{\Gamma (f)}$. So the $(f \circ p_1) \mid _{V_{\Gamma (f)}} \simeq p_2 \mid _{V_{\Gamma (f)}}$. Therefore the cohomology classes $(f \circ p_1)^*(a) = p_1^* \circ f^*(a) = f^*(a) \times 1$ and $p_2^*(a) = 1 \times a$ have the same image under the restriction homomorphism $H^i(M \times N; {\mathbb {Z}}) \rightarrow H^i(V_{\Gamma (f)}; {\mathbb {Z}})$. Now using the commutative diagram

it follows that $(f^*(a) \times 1) \cup \tilde{\mu } = (1 \times a) \cup \tilde{\mu }.$ So $j^*((f^*(a) \times 1) \cup \tilde{\mu }) = j^*((1 \times a) \cup \tilde{\mu }),\; j$ as defined earlier. By the properties of mixed cup products $(f^*(a) \times 1) \cup j^*(\tilde{\mu }) = (1 \times a) \cup j^*(\tilde{\mu }).$ So $(f^*(a) \times 1) \cup \tilde{\mu }' = (1 \times a) \cup \tilde{\mu }'.$ $\square$

The next theorem is an analogue of Theorem 11.11 of [15] (and also of Proposition (30.18) of [6]):

Theorem 2.4

Let $\{b_i\}^{l_N}_{i=1}$ be a homogeneous basis of $H^*(N; {\mathbb {Q}})$, and let $\{b_i^{*}\}^{l_N}_{i=1}$ be its Poincaré dual basis, then the cohomology class $\tilde{\mu }'$ is given by

$$\begin{aligned} \tilde{\mu }' = \sum _{i=1}^{l_N} (-1)^{\deg \, (b_i)}f^*(b_i) \times b_i^{*}, \end{aligned}$$

where deg $(b_i)=k$ if $b_i\in H^k(N; {\mathbb {Q}}).$

Proof

Using the Künneth formula we can write $\tilde{\mu }' \in H^n(M \times N ; {\mathbb {Q}})$ as

$$\begin{aligned} \tilde{\mu }' = c_1 \times b_1^{*} +...+c_{l_N} \times b^{*}_{l_N}, \end{aligned}$$

where $c_1,\ldots , c_{l_N}$ are certain well defined cohomology classes in $H^*(M; {\mathbb {Q}})$ with $\deg \, (b_i) + \deg \, (c_i)= n$. We apply the homomorphism $/[N]$ on both sides of the identity $(f^*(a) \times 1) \cup \tilde{\mu }' = (1 \times a) \cup \tilde{\mu }',$ where $a \in H^*(N; {\mathbb {Q}})$. By Property 4 of slant product on p. 288 of Spanier [19] and by our Theorem 2.2 we get

$$\begin{aligned} ((f^*(a) \times 1) \cup \tilde{\mu }') / [N] = f^*(a) \cup (\tilde{\mu }'/ [N]) = f^*(a). \end{aligned}$$

(1)

Also,

$$\begin{aligned} ((1 \times a) \cup \tilde{\mu }')/[N]&= (1 \times a) \cup \sum _{i=1}^{l_N} c_i \times b_i^{*} /[N] = \sum _{i=1}^{l_N} (-1)^{\deg \, (a) \deg \, (c_i)} c_i \nonumber \\&\times (a \cup b_i^{*}) /[N] \nonumber \\&= \sum _{i=1}^{l_N} (-1)^{\deg \, (a) \deg \, (c_i)} c_i \langle a \cup b_i^{*}, [N]\rangle \end{aligned}$$

(2)

Using equations (1) and (2) above and substituting $b_j$ for $a$, we get

$$\begin{aligned} f^*(b_j) = \sum _{i=1}^{l_N} (-1)^{\deg \, (b_j) \deg \, (c_i)} c_i \langle b_j \cup b_i^{*}, [N]\rangle = (-1)^{\deg \, (b_j) \deg \, (c_j)} c_j, \end{aligned}$$

but $\deg \, (b_j) = \deg \, (c_j)$, so we get that

$$\begin{aligned} (-1)^{\deg \, (b_j)} f^*(b_j) \!=\! (-1)^{\deg \, (b_j)}(-1)^{\deg \, (b_j) \deg \, (b_j)} c_j \!=\! (-1)^{\deg \, (b_j) (1 + \deg \, (b_j)} c_j = c_j. \end{aligned}$$

This proves the theorem. $\square$

Let $[\Gamma (f)] = (1_M \times f)_* \circ {\Delta _M}_* ([M])$. Let $\mu _N \in H^n(N \times N, N \times N \backslash \Gamma (1_N); {\mathbb {Z}})$ be the Thom class of the oriented normal bundle of the embedding $\Gamma (1_N) \hookrightarrow N \times N$, let $\mu _N' \in H^n(N \times N; {\mathbb {Z}})$ be its image under the homomorphism in cohomology induced by $j_N :N \times N \rightarrow (N \times N, N \times N \backslash \Gamma (1_N))$ and $\mu _f = (f \times 1_N)^* (\mu _N') \in H^n(M \times N; {\mathbb {Z}})$.

Remark 2.5

• 1. When coefficients belong to ${\mathbb {Q}}$ (or any field) by Theorem 11.11 of [15] (or by Proposition (30.18) of [6]), if $\{a_i\}_{i=1}^{l_M}$ is a homogeneous basis of $H^*(M;\mathbb {Q})$ and $\{a_i^*\}_{i=1}^{l_M}$ its Poincaré dual basis, we have $\mu _M' = \sum _{i=1}^{l_M} (-1)^{\deg \, (a_i)} a_i \times a_i^{*}$, and $\mu _N' = \sum _{i=1}^{l_N} (-1)^{\deg \, (b_i)} b_i \times b_i^{*}$. Therefore, we see using Theorem 2.4 , that in rational cohomology, $\mu _f = \tilde{\mu }'$.

• 2. By Exercises (30.15) and (30.17) of [6] we have $[M \times M] \cap \mu _M' = {\Delta _M}_*([M]) = [\Gamma (1_M)]$, and $[N \times N] \cap \mu _N' = {\Delta _N}_*([N]) = [\Gamma (1_N)]$. Using this and Part 1 of this remark one can prove the following statement which is the assertion of the Exercise (30.22) of [6]:

  $$\begin{aligned}{}[M] \times [N] \cap \tilde{\mu }' = [M] \times [N] \cap \mu _f = [\Gamma (f)] \end{aligned}$$

  (A)

We are now ready to prove Theorem 1.8.

Proof of Theorem.1.8

Let $h:M \times N\!\setminus \!\Gamma (f)\hookrightarrow M \times N$ denote the inclusion map. The Leray spectral sequence for this inclusion has the form

$$\begin{aligned} E_2^{i, j} = H^i(M\times N; R^jh_*\mathbb {Z}) \Rightarrow H^{i+j}(M\times N \setminus \Gamma (f);\mathbb {Z}). \end{aligned}$$

Here $R^jh_*\mathbb {Z}$ is the Leray sheaf on $M \times N$ associated with the presheaf $U \rightarrow H^j(U \cap (M \times N\!\setminus \!\Gamma (f));\mathbb {Z})$ where $U$ runs over the family of all open subsets of $M \times N$. The stalk of $R^jh_*\mathbb {Z}$ at a point $\bar{x}=(x_1,x_2) \in M \times N$ is described below.

• 1. If $x_2 \ne f(x_1)$, we can choose small coordinate neighborhoods $U_1$ and $U_2$ of $x_1$ and $x_2$ in $M$ and $N$ respectively, such that $f(U_1)\cap U_2=\emptyset$. If $U = U_1\times U_2$, then $U \subset M\times N\!\setminus \!\Gamma (f)$. Therefore, for every $\bar{x}=(x_1,x_2)\in M\times N, x_2 \ne f(x_1)$, $(R^jh_*\mathbb {Z})_{\bar{x}} = H^j(U \cap (M\times N\!\setminus \!\Gamma (f));\mathbb {Z}) = H^j(U;\mathbb {Z}) = H^j(U_1\times U_2;\mathbb {Z}) = 0$ if $j\ne 0$, and $(R^0h_*\mathbb {Z})_{\bar{x}} = {\mathbb {Z}}$.

• 2. If $x_2 = f(x_1)$, we can choose small coordinate neighborhoods $U_1$ and $U_2$ of $x_1$ and $x_2$ in $M$ and $N$ respectively, such that $f(U_1)\subset U_2$. For $U = U_1 \times U_2$, $(R^jg_*\mathbb {Z})_{\bar{x}} = H^j(U \cap (M\times N\!\setminus \!\Gamma (f));\mathbb {Z}) = H^j(U\!\setminus \!\Gamma (f);\mathbb {Z}) = H^j(U_1 \times U_2 \setminus \Gamma (f) ;\mathbb {Z})$. Let $h_1: U_1 \xrightarrow {\cong } \mathbb {R}^n\; \text{ and } \; h_2: U_2 \xrightarrow {\cong } \mathbb {R}^n$, be the coordinate charts of $U_1 \;\text{ and }\; U_2$ respectively. Consider the composite $\hat{f} = h_2\circ f \circ h_1^{-1}:\mathbb {R}^n \rightarrow \mathbb {R}^n$, then $\Gamma (\hat{f}) = h_1\times h_2 (\Gamma (f)|_{U_1\times U_2})$. Therefore, $(R^jh_*\mathbb {Z})_{\bar{x}}= H^j(U_1 \times U_2\!\setminus \!\Gamma (f);\mathbb {Z}) \cong H^j((\mathbb {R}^n \times \mathbb {R}^n)\!\setminus \!\Gamma (\hat{f});\mathbb {Z})$. But as observed in the introduction $\mathbb {R}^n\times \mathbb {R}^n\!\setminus \!\Gamma (\hat{f})$ is homeomorphic to $F(\mathbb {R}^n, 2)$, so for every $\bar{x}=(x_1,x_2)\in M\times N, x_2 = f(x_1)$

  $$\begin{aligned} (R^jh_*\mathbb {Z})_{\bar{x}}= H^j(F(\mathbb {R}^n, 2); \mathbb {Z}) = \left\{ \begin{array}{ll} \mathbb {Z}&{} \quad \text{ if } \ j = 0, n-1\\ 0 &{} \quad \text{ otherwise } \end{array} \right. \end{aligned}$$

This shows that the Leray sheaf $(R^{n-1}h_*\mathbb {Z})$ is supported and is locally constant along the graph $\Gamma (f)$ with stalks ${\mathbb {Z}}$, and $(R^0h_*\mathbb {Z})$ is locally constant on $M \times N$ with stalks ${\mathbb {Z}}$. Since $M$, $N$ and hence $\Gamma (f)$ are orientable, we have

$$\begin{aligned} H^{i}(M \times N; R^jh_*\mathbb {Z})= \left\{ \begin{array}{ll} H^{i}(M \times N; \mathbb {Z}) &{} \quad \text{ if } \ j = 0,\\ H^{i}(\Gamma (f); \mathbb {Z}) &{} \quad \text{ if }\ j = n-1,\\ 0 &{} \quad \text{ otherwise } \end{array} \right. \end{aligned}$$

Hence, the $E_2$ terms of Leray spectral sequence take the form:

$$\begin{aligned} \begin{array}{ccccc} 0&{}0&{}\cdots &{}0&{}\cdots \\ E_2^{0,n-1}=H^0(\Gamma (f);\mathbb {Z}) &{} E_2^{1,n-1}=H^1(\Gamma (f);\mathbb {Z}) &{}\cdots &{} E_2^{n,n-1}=H^n(\Gamma (f);\mathbb {Z}) &{}\cdots \\ 0&{}0&{}\cdots &{}0&{}\cdots \\ 0&{}0&{}\cdots &{}0&{}\cdots \\ 0&{}0&{}\cdots &{}0&{}\cdots \\ 0&{}0&{}\cdots &{}0&{}\cdots \\ E_2^{0,0}=H^0(M\times N;\mathbb {Z}) &{} E_2^{1,0}=H^1(M\times N;\mathbb {Z}) &{}\cdots &{} E_2^{n,0}=H^n(M\times N;\mathbb {Z}) &{} \cdots \\ \end{array} \end{aligned}$$

and $E_n$ terms coincide with $E_2$ terms:

We now determine $d_n$ on the $E_n$ page. Note that the differential $d_n$ is 0 on the bottom row, since it maps each row to a lower row. Let $i:\Gamma (f)\rightarrow M\times N$ be the inclusion map and let $\pi :M\times N\rightarrow \Gamma (f)$ be defined by $\pi (x_1,x_2)=(x_1,f(x_1))$. Then $\pi \circ i=1_{\Gamma (f)}$. Therefore, by the functoriality of cohomology, the homomorphisms induced in cohomology, $\pi ^*$ and $i^*$ are, respectively, injective and surjective. Hence the long cohomology exact sequence of the pair $(M \times N, \Gamma (f))$ gives rise to the following split short exact sequence:

$$\begin{aligned} 0 \rightarrow H^*(M \times N, \Gamma (f); {\mathbb {Z}})\xrightarrow {j_1^*} H^*(M \times N; {\mathbb {Z}}) \underset{\pi ^*}{\buildrel {i^*}\over \rightleftharpoons } (H^*(\Gamma (f); {\mathbb {Z}}))\rightarrow 0, \end{aligned}$$

where $j_1:M \times N \rightarrow (M\times N,\Gamma (f))$ is the natural injection.

As $\Gamma (f)$ is a smooth submanifold with an orientable normal bundle in the smooth manifold $M \times N$, the differentials $d_n$ in the above Leray spectral sequence originating from the $(n-1)^{th}$ row are Gysin maps. If $z \in H^i(M \times N; {\mathbb {Z}}), \; i^*(z)\in H^i(\Gamma (f); {\mathbb {Z}})$ and we have $d_n(i^*(z)) = \pi ^*(i^*(z)) \cup \tilde{\mu }'$. From the above split exact sequence we see that $z - \pi ^*(i^*(z)) = j_1^*(w)$, for some element $w \in H^i(M \times N, \Gamma (f); {\mathbb {Z}})$. Recall that $\tilde{\mu }' = j^*(\tilde{\mu })$, the image of the Thom class $\tilde{\mu } \in H^*(M \times N, M \times N \setminus \Gamma (f); {\mathbb {Z}})$, where $j : M \times N \rightarrow (M \times N, M \times N\!\!\setminus \!\!\Gamma (f))$ is the natural injection. So by Property 8 on p. 251 of [19] and the fact that $w \cup \tilde{\mu } = 0$ (since the mixed cup product $\cup : H^{i}(M \times N,\Gamma (f);\mathbb {Z}) \otimes H^{n} (M \times N,M \times N \setminus \Gamma (f);\mathbb {Z}) \rightarrow H^{i+n}(M \times N,\Gamma (f)\cup M \times N \setminus \Gamma (f) ;\mathbb {Z}) = H^{i+n}(M \times N, M \times N;\mathbb {Z}) = 0$ has its image in a zero module) we get that $(z - \pi ^*(i^*(z))) \cup \tilde{\mu }' = j_1^*(w) \cup \tilde{\mu }' = j_1^*(w) \cup j^*(\tilde{\mu }) = 0$. This means, in particular, that $d_n(1) = \tilde{\mu }'$.

If $k: M \rightarrow \Gamma (f)$ denotes the mapping $x \mapsto (x,f(x))$ and $\pi _M:M \times N \rightarrow M$ be the projection onto $M$, then $k^*$ is an isomorphism and $\pi ^*=\pi _M^*\circ k^*$. Thus $d_n(i^*(z)) = \pi _M^*(k^*(i^*(z))) \cup \tilde{\mu }' = (k^*(i^*(z)) \times 1) \cup \tilde{\mu }'$. But by Property 4 of slant products on p. 288 of [19], $\{(k^*(i^*(z)) \times 1)\cup \tilde{\mu }'\}/[N] = (k^*(i^*(z))\cup (\tilde{\mu }'/[N]) = k^*(i^*(z))$, as $\tilde{\mu }'/[N] = 1$ by Theorem 2.2 above. So we have: if $i^*(z)\ne 0$ then $(k^*(i^*(z))\times 1) \cup \tilde{\mu }'\ne 0$. Therefore $d_n$ is injective on the $(n-1)^{th}$-row of $E_n = E_2$.

This yields $E_{n+1}^{i,j} = E_{\infty }^{i,j} = 0$ for $j \ne 0$, and $E^{i,0}_{\infty } = E^{i,0}_{n+1} = \frac{E^{i,0}_{n}}{{{\mathrm{im}}}d_n} = \frac{E^{i,0}_{2}}{{{\mathrm{im}}}d_n} = \frac{H^i(M \times N; {\mathbb {Z}})}{{{\mathrm{im}}}d_n}$ for all $i$. But, for $0 \le i \le n-1$, we have $E_{\infty }^{i,0} = E_{n+1}^{i,0} = E_{n}^{i,0} = E_{2}^{i,0} = H^i(M \times N; {\mathbb {Z}})$, and for $n \le i \le 2n$, $E_{n+1}^{i,0} = E_{\infty }^{i,0} = \frac{H^i(M \times N; {\mathbb {Z}})}{{{\mathrm{im}}}d_n}$.

Thus

$$\begin{aligned} H^*(M \times N\!\!\setminus \!\!\Gamma (f); {\mathbb {Z}}) \cong \frac{H^*(M \times N; {\mathbb {Z}})}{H^*(M \times N; {\mathbb {Z}}) \cup \tilde{\mu }'} \cong \frac{H^*(M \times N; {\mathbb {Z}})}{(\tilde{\mu }')}, \end{aligned}$$

as rings, where $(\tilde{\mu }')$ denotes the ideal of $H^*(M \times N; {\mathbb {Z}})$ generated by $\tilde{\mu }'$.

If we take cohomology with coefficients in $\mathbb {Q}$, then by an application of the Künneth theorem $H^*(M; {\mathbb {Q}}) \otimes H^*(N; {\mathbb {Q}}) \xrightarrow {\cong } H^*(M \times N; {\mathbb {Q}})$. Hence there is a unique class $\Delta _f =\sum ^n_{i=1}(-1)^{\deg \,(b_i)}f^*(b_i)\otimes b_i^{*}$ in $H^*(M; {\mathbb {Q}}) \otimes H^*(N; {\mathbb {Q}})$ which maps to $\tilde{\mu }' = \sum ^N_{i=1}(-1)^{\deg \,(b_i)} f^*(b_i)\times b_i^{*} = \mu _f$ in $H^*(M \times N; {\mathbb {Q}})$ under this isomorphism, and therefore

$$\begin{aligned} H^*(M \times N \setminus \Gamma (f); {\mathbb {Q}}) \cong \frac{H^*(M \times N; {\mathbb {Q}})}{(\mu _f)} \cong \frac{H^*(M; {\mathbb {Q}})\otimes H^*(N; {\mathbb {Q}})}{(\Delta _f)}, \end{aligned}$$

as algebras. $\square$

A similar argument can be given when $\mathbb {Q}$ is replaced by an arbitrary field.

3 A CDGA model of $M \times N \backslash \Gamma (f)$

In this section we prove Theorem 1.4 and Corollary 1.5 by constructing a CDGA model of $(1_M \times f) \circ \Delta _M: M \rightarrow M\times M \rightarrow M \times N$ from a CDGA model of $f: M \rightarrow N$. (We refer the reader to [9], [10], [3] and [5] for necessary definitions, notations and results leading to the construction of a CDGA model of the configuration space of two points $F(M,2)$).

Proof of Theorem 1.4

We first note that a CDGA-model $(A_N, d_N)\xrightarrow {\psi _f} (A_M, d_M)$ of $f$ determines a CDGA-model $(A_M \otimes A_N, d_{A_M} \otimes d_{A_N})$ $\xrightarrow {1_{A_M} \otimes \psi _f}$ $(A_M \otimes A_M, d_{A_M} \otimes d_{A_M})$ of $1_M \times f$, whose verification is left to the reader. This together with the fact that a CDGA-model of $M$ determines a CDGA-model of $\Delta _M$ (see, e.g. Example 2.48 on p.73 of [5]) yields the conclusion that a CDGA-model of $f$ determines a CDGA-model $(A_M \otimes A_N) \xrightarrow {1_{A_M} \otimes \psi _f} (A_M \otimes A_M)\xrightarrow {\nu _{A_M}} A_M$, defined by $x\otimes y \mapsto (\nu _{A_M} \circ (1_{A_M} \otimes \psi _f)) (x\otimes y) = x.\psi _f(y),$ of $(1_M \times f) \circ \Delta _M$.

By our hypotheses $H^1(M;\mathbb {Q}) = 0$ and $H^1(N;\mathbb {Q}) = 0 = H^2(N;\mathbb {Q}),$ it follows that for the embedding $(1_M \times f) \circ \Delta _M : M \rightarrow M \times N$, which actually embeds $M$ as the graph $\Gamma (f)$ of $f$ in $M \times N$, the hypotheses of Theorem 1.4 of [10], namely $H_1((1_M \times f) \circ \Delta _M;\mathbb {Q})$ is an isomorphism and $H_2((1_M \times f) \circ \Delta _M;\mathbb {Q})$ is an epimorphism are satisfied. Therefore by applying Theorem 1.4 of [10] one can determine explicitly a model of $M \times N \setminus \Gamma (f)$ from the above model of $f$, and our result follows. $\square$

Proof of Corollary. 1.5

By hypothesis the dimension of $N$ is at least three, so $\text{ dim } \;M = \text{ dim }\; N \ge 3$. Therefore the hypothesis of Corollary 1.5 of [10], namely the codimension of the embedding of $\Gamma (f)$ in $M \times N$ is $\ge 3$, and $H_*((1_M \times f) \circ \Delta _M;\mathbb {Q})$ is 2-connected are satisfied. Hence, the rational homotopy type of $M \times N \backslash \Gamma (f)$ is determined by the rational homotopy class of $1_M \times f$. But the homotopy class of $1_M \times f$ is determined by the homotopy class of $f$; hence the corollary holds. $\square$

4 The CDGA model of $M \times N \backslash \Gamma (f)$ based on differential-Poincaré-duality-algebra models of $M$ and $N$

In this section we consider a continuous map $f:M \rightarrow N$ of closed connected oriented manifolds of dimension $n$ with $H^1(M; {\mathbb {Q}}) = 0 = H^1(N; {\mathbb {Q}})$ having a CDGA model $(A_N, d_{A_N}) \buildrel {\psi _f}\over \rightarrow (A_M, d_{A_M})$ in which $(A_M, d_{A_M})$ and $(A_N, d_{A_N})$ are differential connected Poincaré duality algebras of formal dimension $n$ (refer to the para preceding Theorem 1.6 of the introduction).

We note that all the algebras and relevant results of [9], proved by them in the absolute case, can analogously be developed for any given CDGA morphism of Poincaré duality algebras.

Definition 4.1

Let $(A_2, d_{A_2}) \buildrel {\psi }\over \rightarrow (A_1, d_{A_1})$ be a morphism of Poincaré duality algebras of formal dimension $n$. Define $\Delta _{\psi } := (\psi \otimes 1)(\Delta ),$ where $\Delta$ is the diagonal class defined in the statement of Theorem 1.6 and the element $\Delta _{\psi }$, which belongs to $(A_1 \otimes A_2)^n$, will be called the class of $\psi$.

As in Proposition 4.3 of [9] the element $\Delta _{\psi }$ does not depend on the choice of the basis $\{a_i\}_{1\le i \le l}$ of $A_2$.

Remark 4.2

If $s^{-n}A_1$ is the suspension of $A_1$ as defined in §2 of [9] then there is an $(A_1 \otimes A_2)$-module structure on $s^{-n}A_1$ given by

$$\begin{aligned} (x \otimes y).(s^{-n}a)=(-1)^{n\deg \,(x)+n\deg \,(y)+\deg \,(a)\deg \,(y)}s^{-n}(x.a.\psi (y)) \end{aligned}$$

for homogeneous elements $a, x \in A_1$, and $y \in A_2$ and an obvious $(A_1 \otimes A_2)$-module structure on $(A_1 \otimes A_2)$. In other words the $(A_1 \otimes A_2)$-module structure on $s^{-n}A_1$ is obtained from the obvious structure of $A_1 \otimes A_1$-module on $s^{-n} A_1$ by transporting it along the CDGA map $1\otimes \psi : A_1 \otimes A_2 \rightarrow A_1 \otimes A_1.$

In view of the above, statements below follow from the corresponding statements of [9], proved there in the absolute case.

• 1. Let $\psi : (A_2, \omega _{A_2}) \rightarrow (A_1, \omega _{A_1})$ be a CDGA morphism of oriented Poincaré duality algebras of finite dimension and of formal dimension $n$. Then the map $\widehat{\Delta _{\psi }}:s^{-n} A_1 \rightarrow A_1 \otimes A_2,\;\text{ defined } \text{ by }\; s^{-n}a \mapsto \Delta _{\psi }.(a \otimes 1)$ is a morphism of ($A_1 \otimes A_2$)-modules.

• 2. Let $\psi : (A_2, d_{A_2}, \omega _{A_2}) \rightarrow (A_1, d_{A_1}, \omega _{A_1})$ be a morphism of oriented differential Poincaré algebras of formal dimension $n$. Then $\Delta _{\psi }$ is a cocycle.

• 3. If $\psi : (A_2, d_{A_2}, \omega _{A_2}) \rightarrow (A_1, d_{A_1}, \omega _{A_1})$ is a morphism of connected differential oriented Poincaré duality algebras of formal dimension $n$ and $\widehat{\Delta _{\psi }}:s^{-n} A_1 \rightarrow A_1 \otimes A_2\;\text{ is } \text{ defined } \text{ by }\; s^{-n}a \mapsto \Delta _{\psi }.(a\otimes 1),$ then

  • (a) $\widehat{\Delta _{\psi }}$ is a morphism of ($A_1 \otimes A_2$)-dgmodules and hence its mapping cone

    $$\begin{aligned} C(\widehat{\Delta _{\psi }}):= A_1 \otimes A_2\oplus _{\widehat{\Delta _{\psi }}}ss^{-n}A_1 \end{aligned}$$

    is an ($A_1 \otimes A_2$)-dgmodule.

  • (b) $C(\widehat{\Delta _{\psi }})$ is also a $CDGA$ under the following multiplication rules: If $a, a', x, y \in A_1$, and $b, b' \in A_2$,

    $$\begin{aligned} (a \otimes b).(a'\otimes b')&= (-1)^{\deg \,(b)\deg \,(a')}(a.a'\otimes b.b'),\\ (a \otimes b).(ss^{-n} x)&= (-1)^{(n-1)(\deg \,(a)+\deg \,(b))}ss^{-n}(a.\psi (b).x), \\ (ss^{-n} x).(a \otimes b)&= ss^{-n}(x.a.\psi (b)),\\ (ss^{-n} x).(ss^{-n} y)&= 0. \end{aligned}$$

Definition 4.3

Let $(A,\omega )$ be a connected oriented Poincaré duality algebra of formal dimension $n$ with orientation class $\omega$ (refer to Definition 4.1 of [9]). Since $A^n \cong \mathbb {Q}$, there exists a unique element $\mu \in A^n$ such that $\omega (\mu )=1$ which called the fundamental class of $A$.

Our goal in this section is to prove Theorem 1.6. Throughout we consider morphisms $\psi : (A_2, d_{A_2}, \omega _{A_2}) \rightarrow (A_1, d_{A_1}, \omega _{A_1})$ of connected differential oriented Poincaré duality algebras of formal dimension $n$.

Lemma 4.4

The ideal $(\Delta _{\psi }):=\Delta _{\psi }.(A_1 \otimes A_2)$ generated by $\Delta _{\psi }$ in $A_1 \otimes A_2$ is a differential ideal and the quotient $(A_1 \otimes A_2)/(\Delta _{\psi })$ is a CDGA.

Proof

By Remark 4.2 (2), $d_{A_1 \otimes A_2}(\Delta _{\psi })=0$; hence the ideal $(\Delta _{\psi })$ is a differential ideal. This implies immediately that the quotient $(A_1 \otimes A_2)/(\Delta _{\psi })$ inherits a $CDGA$ structure.

$\square$

Lemma 4.5

The map $\widehat{\Delta _{\psi }}$ induces an isomorphism

$$\begin{aligned} \widehat{\Delta _{\psi }}:s^{-n}A_1 \rightarrow (\Delta _{\psi }). \end{aligned}$$

Proof

We first show that im $(\widehat{\Delta _{\psi }})=(\Delta _{\psi })$. Clearly $\mathrm{im}(\widehat{\Delta _{\psi }})\subseteq (\Delta _{\psi })$. For the reverse inclusion we have

$$\begin{aligned} (\Delta _{\psi })&= \{\Delta _{\psi }.(x\otimes y)\;|x\in A_1, y\in A_2\}\\&= \{\Delta _{\psi }.(-1)^{\deg \,(x) \deg \,(y)}(1 \otimes y).(x\otimes 1)\;|x\in A_1, y\in A_2\}\\&= \{\widehat{\Delta _{\psi }}((-1)^{\deg \,(x) \deg \,(y)} s^{-n}(\psi (y).x))\;|x\in A_1, y\in A_2\}\\&\subseteq \mathrm{im}(\widehat{\Delta _{\psi }}). \end{aligned}$$

Next we show that $\widehat{\Delta _{\psi }}$ is injective. Let $\omega _{A_1}$ and $\omega _{A_2}$ be the orientation classes of the Poincaré duality algebras $A_1$ and $A_2$ respectively. So there exist unique elements $\mu _{A_1} \in A_1^n$ and $\mu _{A_2} \in A_2^n$, fundamental classes of $A_1$ and $A_2$, such that $\omega _{A_1}(\mu _{A_1}) = 1$ and $\omega _{A_2}(\mu _{A_2}) = 1$. Fix a basis $\{a_i\}_{i=1}^l$ of $A_2$ and its Poincaré dual basis $\{a_i^*\}_{i=1}^l$. Since $A_2^0=\mathbb {Q}$ we can assume that $1 = a_1\in A_2^0$ and remaining $a_i,$’s are of degree $>0$ so that $a_i \mu _{A_2} \in A_2^{> n} = 0$ for $i > 1$. Also $\omega _{A_2}(a_1.a_1^*) = 1$ implies that $1.a_i^* = a_1.a_1^* = \mu _{A_2}$ so that $a_1^*=a_1^{-1}.\mu _{A_2} = \mu _{A_2}$. This yields from the definition of $\Delta _{\psi }$ that

$$\begin{aligned}&\Delta _{\psi }.(\mu _{A_1} \otimes 1)=\sum ^l_{i=1}(-1)^{\deg \,(a_i)}(\psi (a_i) \otimes a_i^*).(\mu _{A_1} \otimes 1)\\&\quad = \sum ^l_{i=1}(-1)^{\deg \,(a_i^*)}(-1)^{n\deg \,(a^*_i)}(\psi (a_i). \mu _{A_1} \otimes a_i^*) = \psi (a_1).\mu _{A_1} \otimes a_1^* \\&\quad = \psi (a_1).\mu _{A_1} \otimes a_1^{-1}.\mu _{A_2} = \mu _{A_1} \otimes \mu _{A_2} \ne 0. \end{aligned}$$

Let $a$ be a non-zero element of $A_1^i$. By Poincaré duality there exists an element $b \in A_1^{n-i}$ such that $a.b=\mu _{A_1}.$ We have

$$\begin{aligned} \widehat{\Delta _{\psi }}(s^{-n} a).(b \otimes 1)&= \Delta _{\psi }.(a\otimes 1).(b \otimes 1) = \Delta _{\psi }.(a.b \otimes 1)= \Delta _{\psi }.(\mu _{A_1} \otimes 1)\\&= \mu _{A_1} \otimes \mu _{A_2} \ne 0. \end{aligned}$$

$\square$

Consider the projection $A_1 \otimes A_2 \xrightarrow {\pi }(A_1 \otimes A_2)/(\Delta _{\psi }).$ We extend $\pi$ to a map $\hat{\pi }: A_1 \otimes A_2 \oplus _{\widehat{\Delta _{\psi }}}ss^{-n} A_1 \rightarrow (A_1 \otimes A_2)/(\Delta _{\psi })$ by setting $\hat{\pi }(ss^{-n} A_1) = 0$.

Lemma 4.6

The map $\hat{\pi }: A_1 \otimes A_2 \oplus _{{\widehat{\Delta _{\psi }}}}ss^{-n} A_1 \rightarrow (A_1 \otimes A_2)/(\Delta _{\psi })$ defined above is a CDGA quasi-isomorphism.

Proof

The map $\hat{\pi }$ can be seen to be a $CDGA$ morphism by a straightforward computation. Since $\widehat{\Delta _{\psi }}$ is injective , we have a short exact sequence

$$\begin{aligned} 0\rightarrow s^{-n}A_1\xrightarrow {{\widehat{\Delta _{\psi }}}} A_1\otimes A_2\xrightarrow {{\pi }}(A_1\otimes A_2)/\mathrm{im}(\widehat{\Delta _{\psi }})\rightarrow 0. \end{aligned}$$

Comparing the long cohomology exact sequence corresponding to this exact sequence and that of the mapping cone $s^{-n} A_1 \xrightarrow {\widehat{\Delta _{\psi }}} A_1\otimes A_2 \hookrightarrow C(\widehat{\Delta _{\psi }}):= A_1\otimes A_2 \oplus _{\widehat{\Delta _{\psi }}}ss^{-n} A_1$ we get the following commutative diagram:

Applying five lemma and using Lemma 4.5 we get that the map

$$\begin{aligned} A_1\otimes A_2 \oplus _{\widehat{\Delta _{\psi }}}ss^{-n}A_1 \xrightarrow {\hat{\pi }=\pi \oplus 0} (A_1\otimes A_2)/\mathrm{im}(\widehat{\Delta _{\psi }})=(A_1 \otimes A_2)/(\Delta _{\psi }) \end{aligned}$$

is a quasi isomorphism. $\square$

We are now ready to prove Theorem 1.6.

Proof of Theorem. 1.6

We have proved in Lemma 4.4 that $(\Delta _{\psi _f})$ is a differential ideal. Since $(A_M, d_{A_M})$ and $(A_N, d_{A_N})$ are connected differential Poincaré duality algebras of formal dimension $n$ and since $H^n(A_M, d_{A_M})=H^n(M;\mathbb {Q})\ne 0$ and $H^n(A_N, d_{A_N})=H^n(N;\mathbb {Q})\ne 0$, Proposition 4.8 of [9] implies that $(A_M, d_{A_M}, \omega _{A_M})$ and $(A_N, d_{A_N}, \omega _{A_N})$ are oriented differential Poincaré duality algebras in the sense of Definition 4.6 of [9] for orientations $\omega _{A_M} \in \#A_M^n$ and $\omega _{A_N}\in \#A_N^n$.

We have proved in §3 that a CDGA model $\psi _f: (A_N, d_{A_N}) \rightarrow (A_M, d_{A_M})$ of $f$ gives a CDGA model

$$\begin{aligned} \Phi := \nu _{A_M} \circ (1_{A_M} \otimes \psi _f): (A_M \otimes A_N, d_{A_N} \otimes d_{A_M}) \rightarrow (A_M, d_{A_M}) \end{aligned}$$

of $(1_M \times f) \circ \Delta : M \rightarrow M \times N,$ and is given by $x \otimes y \mapsto x.\psi _f(y)$. Set $n=2m$. The morphism $\Phi$ induces an obvious ($A_M\otimes A_N$)-dgmodule structure on $A_M$, hence on $s^{-n}\#A_M$. By Proposition 4.7 of [9] $s^{-n}\#A_M = s^{-m}(s^{-m}\#A_M)$ is isomorphic to $s^{-m}A_M$ as an $A_M$-dgmodule, therefore also an isomorphism as an ($A_M\otimes A_N$)-dgmodule.

By Remark 4.2a the map $\widehat{\Delta _{\psi }}:s^{-n}A_M \rightarrow A_M\otimes A_N$ is a map of ($A_M\otimes A_N$)-dgmodules. Moreover, it induces an isomorphism in the cohomology in the top degree, because $\widehat{\Delta _{\psi }}(s^{-n}\mu )= \mu _{A_M} \otimes \mu _{A_N}$ . Thus $\widehat{\Delta _{\psi }}$ is a shriek map (or, a top-degree map) in the sense of Definition 5.1 of [10].

Let $I_0$ be a complement of the cocycles in $(A_M \otimes A_N)^{n-3}$ and set $I = I_0 \oplus (A_M\otimes A_N)^{>n-3}$. Let $K_0$ be a complement of the cocycles in $(s^{-n}A_M)^{n-2}$ and set $K = K_0 \oplus (s^{-n}A_M)^{>n-2}$. Consider the quotient

$$\begin{aligned} \frac{A_M\otimes A_N \oplus _{\widehat{\Delta _{\psi }}} ss^{-n}A_M}{I \oplus sK}. \end{aligned}$$

By the hypothesis of the theorem concerning $f$, it follows that the embedding $(1_M \times f) \circ \Delta _M: M \hookrightarrow M \times N$, which embeds $M$ as $\Gamma (f)$ in $M \times N$, satisfies the hypothesis of Theorem 1.4 of [10], namely that $H_1((1_M \times f)\circ \Delta _M; {\mathbb {Q}})$ is an isomorphism and $H_2((1_M \times f) \circ \Delta _M; {\mathbb {Q}})$ is an epimorphism, hence by Theorem 1.4 of [10] we get that

$$\begin{aligned} \left( \frac{A_M\otimes A_N \oplus _{\widehat{\Delta _{\psi }}} ss^{-n}A_M}{I \oplus sK}, \bar{d}\right) \end{aligned}$$

is a CDGA model of $M \times N\setminus \Gamma (f)$.

Comparing the semi-trivial CDGA structure as defined in Definition 4.1 of [10] on $(A_M\otimes A_N \oplus _{\widehat{\Delta _{\psi }}}ss^{-n}A_M)/(I \oplus sK)$ with the multiplication in the mapping cone given in this section, we conclude that the projection

$$\begin{aligned} (A_M\otimes A_N \oplus _{\widehat{\Delta _{\psi }}}ss^{-n}A_M, d) \rightarrow \left( \frac{A_M\otimes A_N \oplus _{\widehat{\Delta _{\psi }}}ss^{-n}A_M}{I\oplus sK}, \bar{d}\right) \end{aligned}$$

is a CDGA map. Moreover it is a quasi-isomorphism by Lemma 8.6 of [10].

Therefore $( A_M\otimes A_N \oplus _{\widehat{\Delta _{\psi }}}ss^{-n}A_M, d)$ is a CDGA model of $M\times N \setminus \Gamma (f)$. Hence by Lemma 4.6 $\left( (A_M\otimes A_N)/(\Delta _{\psi }), \overline{d_{A_M} \otimes d_{A_N}}\right)$ is a CDGA model of $M\times N \setminus \Gamma (f)$. $\square$

We finally prove Corollary 1.7 on formality of the complement of the graph of a map.

Proof of corollary. 1.7

Since $f:M \rightarrow N$ is formal, $(H^*(N; {\mathbb {Q}}), 0)\buildrel {f^*}\over \rightarrow . (H^*(M; {\mathbb {Q}}), 0)$ is a model of $f$.

Moreover, since $(H^*(N; {\mathbb {Q}}), 0)$ and $(H^*(M; {\mathbb {Q}}), 0)$ are oriented differential Poincaré duality algebras, by Theorems 1.6 and 1.8,

$$\begin{aligned} \left( \frac{H^*(M;\mathbb {Q})\otimes H^*(N;\mathbb {Q})}{(\Delta _f)}, 0\right) \cong (H^*(M\times N\setminus \Gamma (f);\mathbb {Q}), 0) \;\; \text{(as } \text{ CDGAs) } \end{aligned}$$

is a CDGA-model of $M \times N\setminus \Gamma (f)$, and hence it is a formal space. $\square$

5 Examples and applications

In the last section we saw that if $f:M \rightarrow N$ is a formal map of closed connected oriented formal manifolds of dimension $n$ such that $H^1(M;\mathbb {Q}) = 0 = H^1(N;\mathbb {Q}) = H^2(N;\mathbb {Q}),$ then $M \times N\setminus \Gamma (f)$ is formal (Corollary 1.7). In Proposition 6.6 of [9] Lambrechts and Stanley proved that if $M$ is a closed connected orientable manifold of dimension $n$ such that $M\times M\setminus \Gamma (1_M) = F(M,2)$ is formal then $M$ is formal. We ask a similar question (this is the converse of Corollary 1.7):

Question 5.1

Given closed connected oriented formal manifolds $M$ and $N$ of dimension $n$ and a continuous map $f:M\rightarrow N$ such that $M \times N\setminus \Gamma (f)$ is formal, is $f$ necessarily a formal map?

The answer is in the negative. Here we construct such an example.

Example 5.2

Consider $S^7\xrightarrow {f=(h,c)} S^4\times S^3,$ where $h:S^7\rightarrow S^4$ is the Hopf map and $c:S^7\rightarrow S^3$ is a constant map. It follows from Proposition 2.5 of [1] that any formal map $S^7 \rightarrow S^4\times S^3$ must be nullhomotopic; but $f$ is not nullhomotopic. Therefore $f$ is not formal.

Now we show that $S^7 \times (S^4\times S^3)\setminus \Gamma (f)$ is formal. Since $S^3\xrightarrow {j} S^7\xrightarrow {h} S^4$ is a Hopf fibration we have the following commutative diagram [see example (2.68) page 82 of [5] or Chapter 15 of [3]

Here deg $(x)=4$, deg $(y)=7$, and deg $(z)=3$; $d'x=0$, $d'y=x^2$, $d x=0$, $d y=x^2$, $d z = x;$ and $d''z=0;$ the morphism $\rho :(\Lambda (x,y),d')\rightarrow A_{PL}(S^4)$ is the minimal model of $S^4$, $\sigma :(\Lambda (x,y)\otimes \Lambda (z),d)\rightarrow A_{PL}(S^7)$ is a quasi-isomorphism, $\xi :(\Lambda (z),d'')\rightarrow A_{PL}(S^3)$ is the minimal model of $S^3$ , $i:(\Lambda (x,y),d')\rightarrow (\Lambda (x,y)\otimes \Lambda (z),d)$ is a relative minimal CDGA and $\eta : (\Lambda (x,y)\otimes \Lambda (z),d) \rightarrow (\Lambda (z),d'')$ is the projection.

From the above diagram we get the following commutative diagram

where $i(x)=x,\; i(y)=y\;$ and $i(z)=0$.

Let us define CDGAs

$$\begin{aligned} (B,0):=(\Lambda (e_4,e_3)/I_B, 0), \end{aligned}$$

where $\deg \;(e_4)=4;\;\deg (e_3)=3$, $I_B$ is the ideal generated by $e_4^2,$ and

$$\begin{aligned} (A,d_A):=(\Lambda (u_4,u_3)/I_A, d_A), \end{aligned}$$

where $\deg \;(u_4)=4;\;\deg (u_3)=3,\; I_A$ is the ideal generated by $u_4^2, \; d_A\overline{u_3}=\overline{u_4},$ and $d_A\overline{u_4}=0$. It is easily checked that $(B,0)$, and $(A,d_A)$ are oriented differential Poincaré duality algebras.

We define morphisms

$$\begin{aligned} \rho _B: (\Lambda (x,y)\otimes \Lambda (z),d'\otimes d'')\rightarrow (B, 0), \end{aligned}$$

where $\rho _B(x)=\overline{e_4},\;\rho _B(y)=0,\;\rho _B(z)=\overline{e_3}$ and

$$\begin{aligned} \rho _A: (\Lambda (x,y)\otimes \Lambda (z),d)\rightarrow (A, d_A), \end{aligned}$$

where $\rho _A(x)= \overline{u_4},\;\rho _A(y)=0,\;\rho _A(z)=\overline{u_3}.$

Clearly $\rho _B$ and $\rho _A$ are quasi-isomorphisms.

Define $\psi _f:(B,0)\rightarrow (A,d_A)$ by $\psi _f(\overline{e_4})= \overline{u_4},\;\psi _f(\overline{e_3})= 0$ so that the following diagram is commutative.

Using the fact that if $X\xrightarrow {(k,l)}Y\times Z$ is a continuous map then $A_{PL}(k)\cdot A_{PL}(l): A_{PL}(Y)\otimes A_{PL}(Z) \rightarrow A_{PL}(X)$ is its model, we get the following commutative diagram

Therefore $\psi _f$ is a model of $f=(h,c)$. So, by Theorem 1.6

$$\begin{aligned} \left( \frac{A\otimes B}{(\bigtriangleup _{\psi _f})}, \overline{d_A \otimes 0}\right) \end{aligned}$$

is a CGDA model of $S^7 \times (S^4\times S^3)\setminus \Gamma (f)$.

Note that $\{a_1=1,\; a_2=\overline{e_3},\; a_3=\overline{e_4},\; a_4=\overline{e_4}\cdot \overline{e_3}\}$ is a homogeneous basis of $(B,0)$ with Poincaré dual basis $\{a_1^*=\overline{e_4}\cdot \overline{e_3},\; a_2^*=\overline{e_4},\; a_3^*=\overline{e_3},\; a_4^*=1\}$, therefore by definition

$$\begin{aligned} \bigtriangleup _{\psi _f}&= \psi _f(1)\otimes \overline{e_4}\cdot \overline{e_3}-\psi _f(\overline{e_3})\otimes \overline{e_4}+\psi _f(\overline{e_4})\otimes \overline{e_3} - \psi _f(\overline{e_4}\cdot \overline{e_3})\otimes 1\\&= 1\otimes \overline{e_4}\cdot \overline{e_3}-\overline{u_4}\otimes \overline{e_3}. \end{aligned}$$

Therefore as a CDGA

$$\begin{aligned} \left( \frac{A\otimes B}{(\bigtriangleup _{\psi _f})}, \overline{d_A \otimes 0} = \overline{d}\right) =\;\;\; \mathbb {Q}(\overline{1\otimes 1})\xrightarrow {0} 0 \xrightarrow {0} 0 \xrightarrow {0} \mathbb {Q}(\overline{1\otimes \overline{e_3}}, \;\overline{\overline{u_3} \otimes 1}) \end{aligned}$$

$$\begin{aligned} \xrightarrow {\overline{d}} \mathbb {Q}(\overline{1\otimes \overline{e_4}}, \; \overline{\overline{u_4} \otimes 1}) \xrightarrow {0} 0 \xrightarrow {0}\mathbb {Q}(\overline{\overline{u_3} \otimes \overline{e_3}})\xrightarrow {\overline{d}}\mathbb {Q}(\overline{\overline{u_3}\otimes \overline{e_4}},\;\overline{\overline{u_4}\otimes \overline{e_3}},\; \overline{\overline{u_4}. \overline{u_3} \otimes 1}) \end{aligned}$$

$$\begin{aligned} \xrightarrow {\overline{d}} \mathbb {Q}(\overline{\overline{u_4} \otimes \overline{e_4}}) \xrightarrow {0} 0 \xrightarrow {0} \mathbb {Q}(\overline{\overline{u_4}. \overline{u_3}\otimes \overline{e_3}}) \xrightarrow {0} \mathbb {Q}(\overline{\overline{u_4}. \overline{u_3} \otimes \overline{e_4}}) \xrightarrow {0} 0\cdots \end{aligned}$$

Let $c_3\in H^3(S^3;\mathbb {Q})$, $c_4\in H^4(S^4;\mathbb {Q})$ and $c_7\in H^7(S^7;\mathbb {Q})$ be bases of the respective vector spaces. Then $\{b_1=1 \otimes 1,\; b_2=1 \otimes c_3,\; b_3=c_4 \otimes 1,\;b_4 = c_4 \otimes c_3\}$ form a homogeneous basis of $H^*(S^4\times S^3; \mathbb {Q}) = H^*(S^4; \mathbb {Q}) \otimes H^*(S^3; \mathbb {Q})$ with Poincaré dual basis $\{b_1^* = c_4 \otimes c_3,\; b_2^*=c_4 \otimes 1,\; b_3^*=1 \otimes c_3,\;b_4^*=1 \otimes 1\}.$ We know by Theorem 1.8 that

$$\begin{aligned} H^*(S^7 \times (S^4\times S^3)\!\setminus \! \Gamma (f);\mathbb {Q})=\frac{H^*(S^7;\mathbb {Q})\otimes H^*(S^4\times S^3;\mathbb {Q})}{( \bigtriangleup _f)}, \end{aligned}$$

where

$$\begin{aligned} \bigtriangleup _f&= f^*(1 \otimes 1)\otimes c_4\otimes c_3 - f^*(1 \otimes c_3)\otimes c_4 \otimes 1 + f^*(c_4 \otimes 1)\otimes 1 \otimes c_3 \\&- f^*(c_4\otimes c_3)\otimes 1\otimes 1\\&= 1\otimes c_4\otimes c_3. \end{aligned}$$

Therefore as a CDGA

$$\begin{aligned}&(H^*(S^7 \times (S^4\times S^3)\!\setminus \! \Gamma (f);\mathbb {Q}),\; 0) \\&=\;\mathbb {Q}(\overline{1 \otimes 1 \otimes 1})\xrightarrow {0}0\xrightarrow {0}0\xrightarrow {0}\\&\mathbb {Q}(\overline{1 \otimes 1\otimes c_3}) \\&\quad \xrightarrow {0} \mathbb {Q}(\overline{1\otimes c_4 \otimes 1})\xrightarrow {0}0\xrightarrow {0}0\xrightarrow {0}\mathbb {Q}(\overline{c_7\otimes 1 \otimes 1})\xrightarrow {0}0\xrightarrow {0}0 \\&\quad \xrightarrow {0} \mathbb {Q}(\overline{c_7\otimes 1 \otimes c_3})\xrightarrow {0}\mathbb {Q}(\overline{c_7\otimes c_4 \otimes 1})\xrightarrow {0}0\ldots \end{aligned}$$

Now we define a CDGA morphism

$$\begin{aligned} \eta :(H^*(S^7 \times (S^4\times S^3)\!\setminus \! \Gamma (f);\mathbb {Q}),\; 0) \rightarrow \left( \frac{A\otimes B}{(\bigtriangleup \psi _f)}, d_A \otimes 0\right) , \end{aligned}$$

by defining it on generators as follows:

$$\begin{aligned} \eta (\overline{1\otimes 1\otimes 1})&= \overline{1\otimes 1}, \eta (\overline{1 \otimes 1 \otimes c_3}) = \overline{1 \otimes \overline{e_3}}, \eta (\overline{1 \otimes c_4 \otimes 1}) = \overline{1\otimes \overline{e_4}} + \overline{\overline{u_4} \otimes 1}, \\ \eta (\overline{c_7 \otimes 1 \otimes 1})&= \overline{\overline{u_4}. \overline{u_3} \otimes 1}, \; \eta (\overline{c_7 \otimes 1 \otimes c_3}) = \overline{\overline{u_4}. \overline{u_3} \otimes \overline{e_3}}, \; \eta (\overline{c_7 \otimes c_4 \otimes 1}) \\&= \overline{\overline{u_4}. \overline{u_3} \otimes \overline{e_4}}. \end{aligned}$$

It is checked easily that $\eta$ is a quasi-isomorphism. Therefore $(H^*(S^7 \times (S^4\times S^3)\!\setminus \! \Gamma (f);\mathbb {Q}),\; 0)$ is a CDGA model of $S^7 \times (S^4\times S^3)\setminus \Gamma (f)$. Hence $S^7 \times (S^4\times S^3)\setminus \Gamma (f)$ is a formal space.

Remark 5.3

It is known (see e.g. Lemma 6.3 of [9]) that if $M$ is a simply connected closed manifold, if $x\in M$, and if $M\!\setminus \! \{x\}$ is formal, then $M$ is formal.

Against this background we may ask the following question.

Question 5.4

Let $f:M\rightarrow N$ be a continuous map between simply connected formal manifolds and let $x\in M$. Suppose that $f|_{M\setminus \{x\}} :M\!\!\setminus \!\!\{x\} \longrightarrow N\!\!\setminus \!\! \{f(x)\}$ is formal. Is it necessarily true that $f$ is formal ?

The following example shows that the answer to this question is in the negative.

Example 5.5

Take $M=S^3$, $N=S^2$ and $f$ the Hopf map $h:S^3\rightarrow S^2$. Now $S^3\!\setminus \!\!\{(0,0,0,1)\} \cong \mathbb {R}^3$ and $S^2\!\setminus \!\! \{(0,0,1)\} \cong \mathbb {R}^2$. Since any continuous map form $\mathbb {R}^3 \rightarrow \mathbb {R}^2$ is formal, $h|_{S^3\setminus \{(0,0,0,1)\}}: S^3\!\setminus \!\! \{(0,0,0,1)\} \longrightarrow S^2\!\setminus \! \{(0,0,1)\}$ is formal. But $h$ is not formal.

We end the paper by recording a few simple applications of Theorem 1.6 and Corollary 1.7:

Application 5.6

• 1. Let $M$ and $N$ be two simply connected closed formal manifolds of dimension $n$. For any element $y\in N$, since the constant map $f_y: M\rightarrow N$ defined by $f_y(x) = y$ for all $x\in X$ is formal, by Corollary 1.7, $M \times N\setminus \Gamma (f)$ is formal. But $M \times N\setminus \Gamma (f)= M\times (N\!\setminus \!\{y\})$. Therefore $M\times (N\!\setminus \!\{y\})$ is formal.

• 2. Let $M$ be a simply connected closed formal manifold of dimension $n$. Since the identity map $f=1_M: M\rightarrow M$ is formal, by Corollary 1.7, $M \times M\!\setminus \! \Gamma (f)$ is formal. But $M \times M\setminus \Gamma (f)=F(M,2)$. Therefore $F(M,2)$ is formal, which is one of the main results (Corollary 6.1) in [9].

• 3. Let $M$ and $N$ be two simply connected Lie groups of dimension $n$. It is known that the minimal model of a Lie group is of the form $(\Lambda V ,0)$. Hence Lie groups are formal and continuous maps between Lie groups are also formal. Therefore, by Corollary 1.7, for a continuous map $f:M\rightarrow N$, $M \times N\!\setminus \! \Gamma (f)$ is formal.