Functorial Factorizations in the Category of Model Categories

Abstract

We prove that any right Quillen functor between arbitrary model categories admits non trivial functorial factorizations that are similar to those of a model structure. Given a monad, operad or a PROP(erad) $\mathcal {O}$, if we apply one of the factorizations to the forgetful functor $\textsf {U}: \mathcal {O}{\text {-Alg}}(\textsf {M}) \longrightarrow \textsf {M}$, we extend the theory of Quillen–Segal $\mathcal {O}$-algebras initiated in Bacard (Higher Struct 4(1):57–114, 2020), without the hypothesis of $\textsf {M}$ being a combinatorial model category.

Proofs

1.1 Proof of Proposition 2.3

Proof

We define the left adjoint $\textsf {E}(\textsf {H},\textsf {K})_*$ as follows. If $[C]= [C^0,C^{1}, \pi _{C}]$ is an object of $(\textsf {M}'\downarrow \textsf {U}')$, then we set $\textsf {E}(\textsf {H},\textsf {K})_*([C])= [\textsf {K}_*C^0, \textsf {H}_*C^{1}, \textsf {K}_*C^0\longrightarrow \textsf {U}(\textsf {H}_*C^{1})]$, where $\textsf {K}_*C^0\longrightarrow \textsf {U}(\textsf {H}_*C^{1})$ is defined as follows. Let $\eta _{C^1}: C^1 \longrightarrow \textsf {H}\textsf {H}_*C^1$ be the unit in the adjunction $(\textsf {H}_*\dashv \textsf {H})$, and set $\alpha _{C} = \textsf {U}'(\eta _{C^1}) \circ \pi _{C}$. Since we have $\textsf {U}'\textsf {H}= \textsf {K}\textsf {U}$, the map $\alpha _{C}$ can be displayed as the composite:

$$\begin{aligned} \alpha _{C}= [C^0 \xrightarrow {\pi _{C}} \textsf {U}'(C^1) \xrightarrow {\textsf {U}'(\eta _{C^1})} \textsf {U}'\textsf {H}(\textsf {H}_*C^1) \xrightarrow {{\text {Id}}} \textsf {K}(\textsf {U}(\textsf {H}_*(C^1)))]. \end{aligned}$$

With the adjunction $(\textsf {K}_*\dashv \textsf {K})$, the map $\alpha _{C}: C^0 \longrightarrow \textsf {K}(\textsf {U}(\textsf {H}_*(C^1)))$ has a unique adjoint-transpose $\alpha _*([C]) : \textsf {K}_*(C^0) \longrightarrow \textsf {U}(\textsf {H}_*C^1)$, which fits in the equality:

$$\begin{aligned} \alpha _{C}= [C^0 \xrightarrow {\eta _{C^0}} \textsf {K}\textsf {K}_*(C^0) \xrightarrow {\textsf {K}(\alpha _*([C]) )} \textsf {K}(\textsf {U}(\textsf {H}_*(C^1)))]. \end{aligned}$$

We have an explicit formula $\alpha _*([C]) = \varepsilon _{\textsf {U}(\textsf {H}_*(C^1))} \circ \textsf {K}_*(\alpha _{C})$, where $\varepsilon : \textsf {K}_*\textsf {K}\longrightarrow {\text {Id}}_{\textsf {M}}$ is the counit in the adjunction $(\textsf {K}_*\dashv \textsf {K})$. In other words, $\alpha _*([C])$ is the composite:

$$\begin{aligned} \textsf {K}_*(C^0) \xrightarrow {\textsf {K}_*(\alpha _{C})} \textsf {K}_*\textsf {K}(\textsf {U}(\textsf {H}_*(C^1))) \xrightarrow {\varepsilon _{\textsf {U}(\textsf {H}_*(C^1))}} \textsf {U}(\textsf {H}_*(C^1)). \end{aligned}$$

We define $\textsf {E}(\textsf {H},\textsf {K})_*([C])= [\textsf {K}_*C^0, \textsf {H}_*C^{1},\alpha _*([C]) ]$. By construction, we have a commutative diagram:

(A.1.1)

This last commutative diagram represents the unit of the adjunction we are about to establish. If $\theta :[C]\xrightarrow {[\theta ^0, \theta ^1] } [D]$ is a morphism in $(\textsf {M}'\downarrow \textsf {U}')$, the map $\textsf {E}_*(\textsf {H},\textsf {K})(\theta ): \textsf {E}_*(\textsf {H},\textsf {K})([C]) \longrightarrow \textsf {E}_*(\textsf {H},\textsf {K})([D])$ is given by the pair $[\textsf {K}_*(\theta ^0), \textsf {H}_*(\theta ^1)]$. We have displayed below this map as the commutative diagram on the right. The diagram on the left commutes since the assignment $[C]\mapsto \alpha _{C}$ is functorial:

These data clearly define a functor $\textsf {E}(\textsf {H},\textsf {K})_*: (\textsf {M}'\downarrow \textsf {U}') \longrightarrow (\textsf {M}\downarrow \textsf {U})$. It remains to establish that we have some functorial isomorphisms of hom-sets:

$$\begin{aligned} {\text {Hom}}(\textsf {E}(\textsf {H},\textsf {K})_*([C]), [X]) \cong {\text {Hom}}([C], \textsf {E}(\textsf {H},\textsf {K})([X])). \end{aligned}$$

To prove this, assume that we have two maps $\sigma ^{0}: \textsf {K}_*(C^0) \longrightarrow X^0$ and $\sigma ^{1}: \textsf {H}_*(C^1) \longrightarrow X^1$ such that the pair $[\sigma ^{0}, \sigma ^{1}]$ defines a morphism $\sigma : \textsf {E}(\textsf {H},\textsf {K})_*([C]) \xrightarrow {[\sigma ^{0}, \sigma ^{1}]} [X]$. By definition, this means that we have a commutative diagram:

(A.1.2)

If we apply the functor $\textsf {K}$ to the last diagram, we get a commutative diagram to which we have concatenated the diagram (A.1.1):

(A.1.3)

This last diagram defines a map $[C]\longrightarrow \textsf {E}(\textsf {H},\textsf {K})([X])$, defined by the pair $[\textsf {K}(\sigma ^{0}) \circ \eta _{C^0}, \textsf {H}(\sigma ^{1}) \circ \eta _{C^1}] = [\varphi _0(\sigma ^{0}), \varphi _1(\sigma ^{1})]$. Here, $\varphi _0$ and $\varphi _1$ are the isomorphisms of hom-sets:

$$\begin{aligned}&\varphi _0: {\text {Hom}}(\textsf {K}_*(C^0), X^0) \xrightarrow {\cong } {\text {Hom}}(C^0, \textsf {K}(X^0)) \\&\varphi _1: {\text {Hom}}(\textsf {H}_*(C^1), X^1) \xrightarrow {\cong } {\text {Hom}}(C^1, \textsf {H}(X^1)). \end{aligned}$$

Conversely, assume that we are given a morphism $\theta : [C]\xrightarrow {[\theta ^0, \theta ^1]} \textsf {E}(\textsf {H},\textsf {K})([X])$, where $\theta ^0 \in {\text {Hom}}(C^0, \textsf {K}(X^0))$ and $\theta ^1 \in {\text {Hom}}(C^1, \textsf {H}(X^1))$. If we set $\sigma ^{0}=\varphi _0^{-1}(\theta ^0)$ and $\sigma ^{1}=\varphi _1^{-1}(\theta ^1)$, then the map $\theta$ is displayed by a commutative diagram identical to the perimeter of (A.1.3). However, in the latter diagram, the upper inner square involving $\textsf {K}((\alpha _*[C]))$ and $\textsf {K}(\pi _{X})$ does not commute yet. Using the uniqueness of the adjoint-transpose map in the adjunction $(\textsf {K}_*\dashv \textsf {K})$ with respect to the diagonal map $C^0 \longrightarrow \textsf {K}(\textsf {U}(X^1))$, we see that this square does commute and is equal to the image under $\textsf {K}$ of the commutative square hereafter:

(A.1.4)

The last diagram defines a map $\textsf {E}(\textsf {H},\textsf {K})_*([C]) \longrightarrow [X]$ given by $[\varphi _0^{-1}(\theta ^0), \varphi _1^{-1}(\theta ^1)]$. By the above, the assignment $[\sigma ^{0}, \sigma ^{1}] \mapsto [\varphi _0(\sigma ^{0}), \varphi _1(\sigma ^{1})]$ provides the required isomorphism of hom-sets. $\square$

1.2 Proof of Lemma 4.3

Proof of Lemma 4.3

We will follow the same argument as in [1, Section 5.4] to prove the lemma. For Assertion (1) we start by projecting the lifting problem in $\textsf {A}$ using the functor $\varPi ^{1}$. This gives a lifting problem defined by $\sigma ^{1}$ and $\beta ^{1}$. The map $\sigma ^{1}$ is a trivial cofibration since $\sigma$ is in ${\text {Cof}}_{{\text {L}}}({\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {inj}}}}) \cap {\text {W}}_{{\text {L}}}(\textsf {M}_\textsf {U}[\textsf {A}])$, and $\beta ^{1}$ is a fibration since $\beta$ is in ${\text {Fib}}_{{\text {L}}}({\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {inj}}}})$. The axiom of the model category $\textsf {A}$ gives a solution $s^{1}: Y^1 \longrightarrow P^1$ to this lifting problem. Part of $s^{1}$ being a solution gives an equality $\textsf {U}(\gamma ^{1})= \textsf {U}(\beta ^{1}) \circ \textsf {U}(s^{1})$. Moreover, $[\gamma ]=[\gamma ^0,\gamma ^{1}]$ being a morphism in $\textsf {M}_\textsf {U}[\textsf {A}]$ implies that $\pi _{Q} \circ \gamma ^0 = \textsf {U}(\gamma ^{1}) \circ \pi _{Y}$.

Now consider the map $\textsf {U}(s^{1}) \circ \pi _{Y}\in {\text {Hom}}_{\textsf {M}}(Y^0,\textsf {U}(P^1))$ and the map $\gamma ^0 \in {\text {Hom}}_{\textsf {M}}(Y^0,Q^0)$. Then by the above, it is not hard to see that these maps complete the pullback data

$$\begin{aligned} \textsf {U}(P^1) \xrightarrow {\textsf {U}(\beta ^{1})} \textsf {U}(Q^1) \xleftarrow {\pi _{Q}} Q^0, \end{aligned}$$

into a commutative square ($\pi _{Q} \circ \gamma ^0= \textsf {U}(\beta ^{1}) \circ \textsf {U}(s^{1}) \circ \pi _{Y}$). By the universal property of the pullback square, there is a unique map $\zeta :Y^0 \longrightarrow \textsf {U}(P^1) \times _{\textsf {U}(Q^1)} Q^0$, making everything compatible. In particular $\gamma ^0$ and $\textsf {U}(s^{1}) \circ \pi _{Y}$ factor through $\zeta$. The original lifting problem in $\textsf {M}_\textsf {U}[\textsf {A}]$ defined by $[\sigma ]$ and $[\beta ]$ is represented by a commutative cube in $\textsf {M}$. If we unfold it, we find that everything commutes in the diagram hereafter:

By inspection, we get a commutative square that corresponds to a lifting problem defined by the map $\sigma ^0:X^0 \longrightarrow Y^0$ and the map $\delta : P^0 \longrightarrow \textsf {U}(P^1) \times _{\textsf {U}(Q^1)} Q^0$:

Now, it suffices to observe that a solution to this lifting problem gives a solution to the original lifting problem. Indeed, if $s^0 : Y^0 \longrightarrow P^0$ is a solution to the last lifting problem, then $s=[s^0,s^{1}]: [Y]\longrightarrow [P]$ is a solution to the original lifting problem. Finally, it is clear that the last lifting problem defined by $\sigma ^0$ and $\delta$ has a solution $s^0 \in {\text {Hom}}_{\textsf {M}}(Y^0,P^0)$ since $\sigma ^0$ is a cofibration and $\delta$ is a trivial fibration. This gives Assertion (1).

For Assertion (2) we proceed as follows. Let $\sigma =[\sigma ^0,\sigma ^{1}]: [X]\longrightarrow [Y]$ be a map in $\textsf {M}_\textsf {U}[\textsf {A}]$. We can use the axiom of the model category $\textsf {A}$ to factor $\sigma ^{1}:X^1 \longrightarrow Y^1$ as a trivial cofibration followed by a fibration, i.e., $\sigma ^{1}= r(\sigma ^{1}) \circ l(\sigma ^{1})$:

The image under $\textsf {U}$ of this factorization, gives a factorization $\textsf {U}(\sigma ^{1})= \textsf {U}(r(\sigma ^{1})) \circ \textsf {U}(l(\sigma ^{1}))$. The map $\textsf {U}(r(\sigma ^{1}))$ is a fibration in $\textsf {M}$ since $\textsf {U}$ preserves fibrations. Let us now consider the pullback square in $\textsf {M}$ defined by the pullback data:

$$\begin{aligned} \textsf {U}(E^{1}) \xrightarrow {\textsf {U}(r(\sigma ^{1}))} \textsf {U}(Y^1) \xleftarrow {\pi _{Y}} Y^0, \end{aligned}$$

and let $p^{1} : \textsf {U}(E^{1}) \times _{\textsf {U}(Y^1)} Y^0 \longrightarrow Y^0$ and $p_2: \textsf {U}(E^{1}) \times _{\textsf {U}(Y^1)} Y^0 \longrightarrow \textsf {U}(E^{1})$ be the canonical maps. Then $p^{1}$ is a fibration in $\textsf {M}$ since the class of fibrations is closed under pullbacks. The universal property of the pullback square gives a unique map $\delta : X^0 \longrightarrow \textsf {U}(E^{1}) \times _{\textsf {U}(Y^1)} Y^0,$ such that everything below commutes.

We can use the axiom of the model category $\textsf {M}$ to factor the map $\delta$ as a cofibration followed by a trivial fibration:

We have an object $[E]=[E^0,E^{1},\pi _{E}]$ in $\textsf {M}_\textsf {U}[\textsf {A}]$, where $\pi _{E}=p_2 \circ b(\delta )$. By the above, we have a map $(i : [X]\longrightarrow [E]) \in {\text {Cof}}_{{\text {L}}}({\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {inj}}}}) \cap {\text {W}}_{{\text {L}}}(\textsf {M}_\textsf {U}[\textsf {A}])$ given by the pair $[a(\delta ), l(\sigma ^{1})]$, and a map $(p : [E]\longrightarrow [Y]) \in {\text {Fib}}_{{\text {L}}}({\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {inj}}}})$ given by the pair $[p^{1} \circ b(\delta ), r(\sigma ^{1})]$. Clearly, we have $\sigma = p \circ i$, which gives Assertion (2).$\square$

1.3 Proof of Lemma 4.10

The proof of this lemma is inspired by the work of Renaudin [12]. We will prove the lemma using some intermediate sublemmas. Let $\tau : [X]\longrightarrow [Q]$ be the diagonal map in the lifting problem:

(A.3.1)

Let $\textsf {M}_\textsf {U}[\textsf {A}]_{/[Q]}:=(\textsf {M}_\textsf {U}[\textsf {A}]\downarrow [Q])$ be the over category whose objects are pairs $([E], \alpha )$, where $\alpha : [E]\longrightarrow [Q]$ is a morphism in $\textsf {M}_\textsf {U}[\textsf {A}]$. The morphisms are the obvious ones. This category inherits from ${\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {proj}}}}$ a model structure called the “over model structure” (see [7, 8]). The object $([Q], {\text {Id}}_{[Q]})$ is a terminal object. With the category $\textsf {M}_\textsf {U}[\textsf {A}]_{/[Q]}$, we can consider the under category of the over category: $([X], \tau )/\textsf {M}_\textsf {U}[\textsf {A}]_{/[Q]} := ([X], \tau ) \downarrow \textsf {M}_\textsf {U}[\textsf {A}]_{/[Q]}$. An object of $([X], \tau )/\textsf {M}_\textsf {U}[\textsf {A}]_{/[Q]}$ is a triple $[([C], u),v]$, where $u: [C]\longrightarrow [Q]$ and $v: [X]\longrightarrow [C]$ are such that $u \circ v = \tau$. In simple terms, this is the category of factorizations of the morphism $\tau : [X]\longrightarrow [Q]$.

Sublemma A.1

With the notation above, the following hold.

• 1. We have two objects of $([X], \tau )/\textsf {M}_\textsf {U}[\textsf {A}]_{/[Q]}$: $[([P], \beta ), \theta ]$ and $[([Y], \gamma ), \sigma ]$.

• 2. The following are equivalent.

  • (a) There is a solution to the lifting problem of the lemma.

  • (b) There is a map $s : [([Y], \gamma ), \sigma ] \longrightarrow [([P], \beta ), \theta ]$ in the category $([X], \tau )/\textsf {M}_\textsf {U}[\textsf {A}]_{/[Q]}$, i.e., we have:

    $$\begin{aligned} {\text {Hom}}([([Y], \gamma ), \sigma ], [([P], \beta ), \theta ] ) \ne \emptyset . \end{aligned}$$

Proof

Clear. $\square$

With this sublemma, we will prove Lemma 4.10 by showing that the hom-set ${\text {Hom}}([([Y], \gamma ), \sigma ], [([P], \beta ), \theta ] )$ is nonempty. To prove this, we will work in the model category $([X], \tau )/{\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {proj}}}}_{/[Q]}$ and use an abstract argument involving Lemma A.3 below.

Sublemma A.2

Given the lifting problem, assume that $\sigma$ is a projective cofibration in $\textsf {M}_\textsf {U}[\textsf {A}]$, and that $\beta$ is a projective (= level-wise) fibration. Then the following hold.

• 1. The object $[([P], \beta ), \theta ]$ is fibrant in $([X], \tau )/{\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {proj}}}}_{/[Q]}$.

• 2. The object $[([Y], \gamma ), \sigma ]$ is cofibrant in $([X], \tau )/{\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {proj}}}}_{/[Q]}$.

Proof

The category $([X], \tau )/{\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {proj}}}}_{/[Q]}$ is given the under model structure from the model category ${\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {proj}}}}_{/[Q]}$. By inspection, a fibrant object in $([X], \tau )/{\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {proj}}}}_{/[Q]}$ is an object $[([E],p),r]$ where $p:[E]\twoheadrightarrow [Q]$ is a level-wise fibration, and $r: [X]\longrightarrow [E]$ is any map such that $\tau = p \circ r$. Dually, a cofibrant object in $([X], \tau )/{\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {proj}}}}_{/[Q]}$ is an object [([B], q), i] where $q:[B] \longrightarrow [Q]$ is any map, and $i : [X]\hookrightarrow [B]$ is a projective cofibration such that $\tau = q \circ i$. $\square$

The last ingredient is the following result which is well known in the theory of model categories. We refer the reader to Hirschhorn [7, Ch. 7].

Lemma A.3

Let $\mathcal {D}$ be a model category. If $g: X \longrightarrow Y$ is a weak equivalence between fibrant objects in $\mathcal {D}$ and C is a cofibrant object of $\mathcal {D}$, then g induces an isomorphism of the sets of homotopy classes of maps: $g_*: \pi (C,X) \longrightarrow \pi (C,Y)$. In particular there is a map $C \longrightarrow X$ in $\mathcal {D}$ if and only if there is a map $C \longrightarrow Y$ in $\mathcal {D}$.

Proof of Lemma 4.10

First, we project the lifting problem in the model category $\textsf {A}$ using the functor $\varPi ^{1}:\textsf {M}_\textsf {U}[\textsf {A}]\longrightarrow \textsf {A}$. This gives a lifting problem defined by $\sigma ^{1}$ and $\beta ^{1}$. The latter problem admits a solution $s^{1}: Y^1\longrightarrow P^1$ since $\sigma ^{1}$ is a trivial cofibration and $\beta ^{1}$ is a fibration. With the same reasoning as in the proof of Lemma 4.3, keeping the same notation, we find that everything below commutes.

Let $[E]= [E^0,E^{1}, \pi _{E}]$ be the object defined within the pullback square, where $E^0= \textsf {U}(P^1) \times _{\textsf {U}(Q^1)} Q^0$, $E^{1}= P^1$ and $\pi _{E}=\textsf {U}(\beta ^1)^*(\pi _{Q})$ is the base change of $\pi _{Q}$:

$$\begin{aligned} \textsf {U}(P^1) \times _{\textsf {U}(Q^1)} Q^0 \xrightarrow {\pi _{E}}\textsf {U}(P^1). \end{aligned}$$

As everything commutes in the above cube, we get the following.

• The pullback square defines a map $\chi :[E]\longrightarrow [Q]$ given by the base change of $\textsf {U}(\beta ^{1})$ and $\beta ^{1}$. The maps $\textsf {U}(\beta ^{1})$ and its base change are fibrations in $\textsf {M}$ since $\beta ^{1}$ is a fibration and $\textsf {U}$ preserves fibrations. It follows that $\chi :[E]\longrightarrow [Q]$ is a level-wise fibration. Consequently, the object $([E],\chi )$ is fibrant in over category ${\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {proj}}}}_{/[Q]}$.

• There is a map $\xi : [P]\longrightarrow [E]$ given by the pair $[\delta ,{\text {Id}}_{P^1}]$. By assumption, $\delta$ is a weak equivalence, therefore $\xi : [P]\xrightarrow {\sim } [E]$ is a level-wise weak equivalence in ${\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {proj}}}}$.

• There is also a map $\widetilde{s}: [Y]\longrightarrow [E]$ defined by $\zeta : Y^0 \longrightarrow \textsf {U}(P^1) \times _{\textsf {U}(Q^1)} Q^0$ and the previous solution $s^{1}: Y^1 \longrightarrow P^1$, i.e., $\widetilde{s}= [\zeta , s^1]$.

It is important to note that these various maps fit in the following factorizations of $\tau : [X]\longrightarrow [Q]$:

These factorizations determine two maps in the under category $([X], \tau )/\textsf {M}_\textsf {U}[\textsf {A}]_{/[Q]}$:

• $\xi : [([P], \beta ), \theta ] \xrightarrow {\sim } [([E], \chi ),\xi \circ \theta ]$. This map is a weak equivalence between objects that are fibrant in the under model category $([X], \tau )/{\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {proj}}}}_{/[Q]}$ (see Sublemma A.2).

• $\widetilde{s} : [([Y], \gamma ), \sigma ] \longrightarrow [([E], \chi ),\xi \circ \theta ]$. The source of this map is a cofibrant object in the under model category $([X], \tau )/{\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {proj}}}}_{/[Q]}$ (see Sublemma A.2).

Since $\xi$ is a weak equivalence between fibrant objects, and since $[([Y], \gamma ), \sigma ]$ is a cofibrant object, we know from Lemma A.3 that there is a map $[([Y], \gamma ), \sigma ] \longrightarrow [([E], \chi ),\xi \circ \theta ]$ if and only if there is a map $[([Y], \gamma ), \sigma ] \longrightarrow [([P], \beta ), \theta ]$, that is:

$$\begin{aligned} {\text {Hom}}([([Y], \gamma ), \sigma ],[([E], \chi ),\xi \circ \theta ]) \ne \emptyset \Longleftrightarrow {\text {Hom}}([([Y], \gamma ), \sigma ],[([P], \beta ), \theta ]) \ne \emptyset . \end{aligned}$$

The hom-set ${\text {Hom}}([([Y], \gamma ), \sigma ],[([E], \chi ),\xi \circ \theta ])$ is nonempty since it contains $\widetilde{s}$, therefore the other hom-set is nonempty, and there exists an element $s \in {\text {Hom}}([([Y], \gamma ), \sigma ],[([P], \beta ), \theta ])$ which is automatically a solution to the original lifting problem by Sublemma A.1. $\square$

1.4 Proof of Lemma 4.11

For the sake of clarity, we will put “p.b” inside a commutative square for a pull-back square, and “p.o” for a pushout square.

Proof of Lemma 4.11

Let $\sigma =[\sigma ^0,\sigma ^{1}]: [X]\longrightarrow [Y]$ be a map in $\textsf {M}_\textsf {U}[\textsf {A}]$. We can use the axiom of the model category $\textsf {A}$ to factor $\sigma ^{1}:X^1 \longrightarrow Y^1$ as a trivial cofibration followed by a fibration, i.e., $\sigma ^{1}= r(\sigma ^{1}) \circ l(\sigma ^{1})$:

The image under $\textsf {U}$ of this factorization gives a factorization:

$$\begin{aligned} \textsf {U}(\sigma ^{1})= \textsf {U}(r(\sigma ^{1})) \circ \textsf {U}(l(\sigma ^{1})). \end{aligned}$$

Moreover, the map $\textsf {U}(r(\sigma ^{1}))$ is a fibration in $\textsf {M}$ since $\textsf {U}$ preserves the fibrations. Let ${Q^0}= \textsf {U}(V^{1}) \times _{\textsf {U}(Y^1)} Y^0$ be the pullback-object obtained by forming the pullback square defined by the data: $\textsf {U}(V^{1}) \xrightarrow {\textsf {U}(r(\sigma ^{1}))} \textsf {U}(Y^1) \xleftarrow {\pi _{Y}} Y^0$. Denote by $p^{1} : {Q^0} \longrightarrow Y^0$ and $p_2: {Q^0} \longrightarrow \textsf {U}(V^{1})$ the canonical maps. Then $p^{1}$ is a fibration in $\textsf {M}$ since the class of fibrations is closed under pullbacks. The universal property of the pullback square gives a unique map $\delta : X^0 \longrightarrow {Q^0}$ such that everything below commutes.

Now we can factor the map $\delta$ as a cofibration followed by a trivial fibration:

The map $\pi _{X}\in {\text {Hom}}_{\textsf {M}}(X^0,\textsf {U}(X^1))$ is equivalent to a morphism $\varphi (\pi _{X}) \in {\text {Hom}}_{\textsf {A}}(\textsf {F}X^0,X^1)$, where $\varphi : {\text {Hom}}_{\textsf {M}}(m,\textsf {U}(a)) \xrightarrow {\cong } {\text {Hom}}_{\textsf {A}}(\textsf {F}m, a)$ is the isomorphism of the adjunction $(\textsf {F}\dashv \textsf {U})$. The last commutative diagram in $\textsf {M}$ corresponds by adjointness to the following commutative diagram in $\textsf {A}$:

The map $\textsf {F}(a_\delta )$ is a cofibration in $\textsf {A}$ since $\textsf {F}$ preserves (trivial) cofibrations. Let $D^{1}=X^1 \cup ^{\textsf {F}X^0} \textsf {F}E^0$ be the pushout-object obtained from the pushout data: . The canonical map $X^1 \hookrightarrow D^{1}$ is also a cofibration as the cobase change of the cofibration $\textsf {F}(a_\delta )$. Also, the universal property of the pushout square gives a unique map $D^{1} \longrightarrow V^{1}$ that we can factor as a cofibration followed by a trivial fibration: . Putting it all together: we get a diagram in $\textsf {A}$ in which everything commutes.

It is important to observe that the composite of cofibrations $(X^1 \hookrightarrow D^{1} \hookrightarrow E^1)$ is a weak equivalence by 3-for-2, since $l(\sigma ^{1}):X^1 \xrightarrow {\sim } V^{1}$ and are weak equivalences. It follows that the composite $X^1 \hookrightarrow E^1$ is in fact a trivial cofibration that will be denoted henceforth by $i^{1}$. Furthermore, since $\textsf {U}$ is a right Quillen functor, the image under $\textsf {U}$ of the trivial fibration is a trivial fibration in $\textsf {M}$. Let $T^0= \textsf {U}(E^1) \times _{\textsf {U}(V^{1})} {Q^0}$ be the pullback-object obtained from the pullback data: . The canonical map is a trivial fibration as the base change of the trivial fibration . By adjointness, the last diagram displayed in the category $\textsf {A}$ corresponds to the commutative diagram in $\textsf {M}$ hereafter, where we have omitted the object $\textsf {U}(D^{1})$ for simplicity:

In this last diagram, the map $E^0 \xrightarrow {\pi _E} \textsf {U}(E^1)$ is adjoint to the composite $(\textsf {F}E^0 \longrightarrow D^{1} \longrightarrow E^1)$. The universal property of the pullback gives a unique (dotted) map $\gamma : E^0 \longrightarrow T^0$. It is important to notice that $\gamma$ is a weak equivalence by 3-for-2 since and are weak equivalences. Another important aspect is that “the pullback of the pullback is a pullback”, therefore the map $T^0\longrightarrow \textsf {U}(E^1)$ is a base change of $\pi _{Y}$ along the composite $\textsf {U}(E^1) \longrightarrow \textsf {U}(V^{1}) \longrightarrow \textsf {U}(Y^1)$. Put differently, the commutative square bounded by the objects $T^0, Y^0, \textsf {U}(Y^1)$ and $\textsf {U}(E^1)$ is a pullback square. Let $j^{1}: E^1 \longrightarrow Y^1$ be the composite fibration , and consider the object $[E]=[E^0,E^{1},\pi _{E}] \in \textsf {M}_\textsf {U}[\textsf {A}]$.

• We have a map $i : [X]\longrightarrow [E]$ given by the pair $[a_\delta , i^{1}]$. By construction, i is a projective cofibration in $\textsf {M}_\textsf {U}[\textsf {A}]$. Moreover, $i^{1}$ is a weak equivalence in $\textsf {A}$, which means that i is a $\varPi ^{1}$-equivalence. Thus i is an element of ${\text {Cof}}_{{\text {L}}}({\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {proj}}}}) \cap {\text {W}}_{{\text {L}}}(\textsf {M}_\textsf {U}[\textsf {A}])$.

• We also have a map $j : [E]\longrightarrow [Y]$ given by the pair $[p^{1} \circ b_\delta , j^{1}]$. By construction, j is a level-wise fibration such that the universal map $(E^0 \longrightarrow \textsf {U}(E^{1}) \times _{\textsf {U}(Y^1)} Y^0 ) = (E^0 \xrightarrow {\sim } T^0)$ is a weak equivalence. Thus j is an element of ${\text {Fib}}_{{\text {L}}}({\textsf {M}_{\textsf {U}}[\textsf {A}]_{{\text {proj}}}})$.

Clearly, we have $\sigma = j \circ i$ and the lemma follows. $\square$