Abstract
In this chapter, we introduce the notion of a simplicial complex of \(\mathbb {R}^n\) and that of a polyhedron. A simplicial complex is a finite assembly of simplices and a polyhedron is a topological space that is homeomorphic to some simplicial complex..
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Appendices
Notes
The Lebesgue lemma (cf. Lemma 3.5.2) about the finite open covers of the n-cube was stated by Lebesgue in [66]. It was used by Lebesgue in order to show that \([0,1]^n\) and \([0,1]^m\) are not homeomorphic for \(n \not = m\). The proof of Lebesgue’s lemma given in [66] contains an error as was pointed out by Brouwer in [19]. A corrected proof appeared in [67].
Let \(D^n\) (resp. \(\mathbb {S}^{n-1}\)) denote the closed unit ball (resp. the unit sphere) in the Euclidean space \(\mathbb {R}^n\). The Brouwer fixed point theorem, which says that every continuous map \(f :D^n \rightarrow D^n\) admits at least one fixed point, may be proved by using tools from algebraic topology (see for example [101]) in the following way.
Suppose for contradiction that there exists a continuous map \(f :D^n \rightarrow D^n\) without fixed points (\(n \ge 2\)). Consider the map \(g :D^n \rightarrow S^{n-1}\) that sends every point \(x \in D^n\) to the intersection point of the half-line starting from f(x) and passing through x with the boundary sphere \(\mathbb {S}^{n-1}\) (see Fig. 3.4). Clearly g is continuous. On the other hand, the map g fixes every point belonging to \(\mathbb {S}^{n-1}\), that is, it satisfies
where h is the inclusion map \(\mathbb {S}^{n-1} \rightarrow D^n\) and i is the identity map on \(\mathbb {S}^{n-1}\). On the level of \(n-1\)-dimensional real homology, the maps g, h and i induce linear maps \(g_* :H_{n-1}(D^n) \rightarrow H_{n-1}(\mathbb {S}^{n-1})\), \(h_* :H_{n-1}(\mathbb {S}^{n-1}) \rightarrow H_{n-1}(\mathbb {S}^{n-1})\) and \(i_* = {{\mathrm{Id}}}:H_{n-1}(\mathbb {S}^{n-1}) \rightarrow H_{n-1}(\mathbb {S}^{n-1})\), where \({{\mathrm{Id}}}\) is the identity map on \(H_{n-1}(\mathbb {S}^{n-1})\). Now it follows from (3.6.1) that
which is impossible since \(H_{n-1}(D^n) = 0\) while \(H_{n-1}(\mathbb {S}^{n-1}) \cong \mathbb {R}\not = 0\). Actually, there are many other proofs of the Brouwer fixed point theorem (see [116] and the references therein). The one presented by Milnor in [76] is elementary an especially clever.
Exercises
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3.1
Show that, up to homeomorphism, there are only countably many polyhedra.
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3.2
Let X be a non-empty topological space. Show that X is a polyhedron with \(\dim (X) = 0\) if and only if X is finite and discrete.
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3.3
Show that a polyhedron has only finitely many connected components.
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3.4
A topological space X is called path-connected if, given any two points \(x,y \in X\), there exists a continuous map \(\gamma :[0,1] \rightarrow X\) such that \(\gamma (0) = x\) and \(\gamma (1) = y\). Show that every connected polyhedron is path-connected.
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3.5
Show that neither the Cantor set nor the Hilbert cube are polyhedra.
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3.6
Find a proof of Lemma 3.5.1 that does not require Baire’s theorem. Hint: proceed by contradiction and use induction on the dimension n (consider a suitable translate of one of the hyperplanes for going from n to \(n - 1\)).
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3.7
(Lebesgue’s lemma for closed coverings). Let \(\alpha \) be a finite closed cover of \([0,1]^n\) such that no element of \(\alpha \) meets two opposite faces of \([0,1]^n\). Show that one has \({{\mathrm{ord}}}(\alpha ) \ge n\). Hint: use Proposition 1.6.3 and Lemma 3.5.2.
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3.8
Show that the unit sphere \(\mathbb {S}^{n - 1} \subset \mathbb {R}^n\), defined by
$$ \mathbb {S}^{n - 1} := \{ (x_1,\dots ,x_n) \in \mathbb {R}^n \mid \; \sum _{k = 1}^n x_k^2 = 1 \}, $$satisfies \(\dim (\mathbb {S}^{n - 1}) = n - 1\).
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3.9
Let X be a polyhedron and n a non-negative integer. Show that one has \(\dim (X) \ge n\) if and only if X contains a subset homeomorphic to \([0,1]^ n\).
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3.10
Let \(\Gamma = (V,\Sigma )\) and \(\Gamma ' = (V',\Sigma ')\) be two abstract simplicial complexes. Equip each of the sets V and \(V'\) with some total ordering. Let \(W := V \times V'\) denote the Cartesian product of the sets V and \(V'\). Consider the set \(\Lambda \) consisting of all subsets \(\lambda \subset W\) satisfying the following condition: there exist an integer \(n \ge 0\), vertices \(v_0,v_1,\dots ,v_n \in V\) and \(v_0',v_1',\dots ,v_n' \in V'\) such that \(v_0 \le v_1 \le \dots \le v_n\), \(v_0' \le v_1' \le \dots \le v_n'\), \(\{v_0,v_1,\dots ,v_n\} \in \Sigma \), \(\{v_0',v_1',\dots ,v_n'\} \in \Sigma '\), and
$$ \lambda = \{(v_0,v_0'),(v_1,v_1'), \dots , (v_n,v_n') \}. $$-
(a)
Show that \(\Pi := (W,\Lambda )\) is an abstract simplicial complex.
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(b)
Let P (resp. \(P'\), resp. Q) denote the support of the geometric realization of \(\Gamma \) (resp. \(\Gamma '\), resp. \(\Pi \)). Show that Q is homeomorphic to the topological product \(P \times P'\).
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(a)
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3.11
Let X and Y be polyhedra. Show that the product space \(X \times Y\) is a polyhedron. Hint: use Exercise 3.10.
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Coornaert, M. (2015). Topological Dimension of Polyhedra. In: Topological Dimension and Dynamical Systems. Universitext. Springer, Cham. https://doi.org/10.1007/978-3-319-19794-4_3
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DOI: https://doi.org/10.1007/978-3-319-19794-4_3
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