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This is pretty basic, but I can't find any answers on this site on this already. I want to know why the homology of two spaces $X$ and $Y$ are the same when $X$ deformation retracts/is homotopy equivalent to $Y$. I would appreciate an intuitive explanation and a formal proof.

Specifically I'm working through Hatcher's Algebraic Topology (chapter 2.1) and by homology, I mean both simplicial and singular homology as defined in Hatcher.

I also assume this holds for relative and reduced homologies, but if you can explain why that would be great.

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    $\begingroup$ Do you want us to produce a proof? It's quite lengthy, plus it can be found in your book (pages 110-113). $\endgroup$ – Danu Sep 22 '16 at 11:29
  • $\begingroup$ It's worth noting that this is actually an axiom of a homology theory, a fundamental property of these kinds of systems in algebraic topology. en.wikipedia.org/wiki/Eilenberg%E2%80%93Steenrod_axioms $\endgroup$ – Alfred Yerger Sep 27 '16 at 2:48
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Thanks to @Danu for the reference. I will give an outline of the proof in Hatcher:

Claim: A map $f : X \to Y$ induces a homomorphism $f_*: H_n(X) \to H_n(Y)$.

Proof: First, define an induced homomorphism $f_\# : C_n(X) \to C_n(Y)$ by $(\sigma : \Delta^n \to X) \mapsto (f \circ \sigma : \Delta^n \to Y)$.

Secondly, we claim that $f_* : H_n(X) \to H_n(Y) : [a] \to [f_\#(a)]$ is a homomorphism. We need that $f_\#(a) \in \ker \partial_n$ whenever $a \in \ker \partial_n$ (so the map makes sense) and $f_\#(b) \in \text{Im }\partial_n$ whenever $b \in \text{Im }\partial_n$ (so the map is well defined, independent of the choice of representative $a$). ($\partial$ is the differential function $\partial_n : C_n(X) \to C_{n-1}(X)$ or $\partial_n : C_n(Y) \to C_{n-1}(Y)$.)

We can check $f_\# \partial = \partial f_\#$ by explicit computation and this gives us what we need.

We have two easy results:

1) $(f \circ g)_* = f_* \circ g_*$ and

2) id$_{*X} = \text{id}_{H_n(X)}$ (the induced map of the identity on $X$ is the identity on $H_n(X)$).

Proof of 1): We have $(f \circ g)_\# = f_\# \circ g_\#$ (by associativity of $\circ$) and therefore $(f \circ g)_* = f_* \circ g_*$.

Proof of 2): By definition id$_{\#X}$ is id$_{C_n(X)}$ and so id$_{*X} = \text{id}_{H_n(X)}$.

Theorem: If two maps $f,g : X \to Y$ are homotopic, then they induce the same homomorphism $f_* = g_* : H_n(X) \to H_n(Y)$.

The proof of this is the meat of the problem and is given in Hatcher (p111-113).

Corollary: The maps $f_* : H_n(X) \to H_n(Y)$ induced by a homotopy equivalence $f: X \to Y$ are isomorphisms for all $n$. Proof: If $f$ is a homotopy equivalence then there is a map $g$ such that $f \circ g$ is homotopy equivalent to id$_Y$ and $g \circ f$ is homotopy equivalent to id$_X$.

Then, by the theorem, $(g \circ f)_* = id_{*X}$ and $(f \circ g)_* = id_{*Y}$. Using the two easy results, we get $g_* \circ f_* = id_{H_n(X)}$ and $f_* \circ g_* = id_{H_n(Y)}$.

Note that a deformation retraction is a special case of a homotopy equivalence so I've now also shown that homologies are invariant under deformation retraction.

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    $\begingroup$ it is exactly the proof I gave you a few days ago. I also gave you a proof of the statement you call "Theorem" in your answer, in my own answer it is called "Step.2 (lemma)" (even if some more details may be welcome as pointed out by Najib Idrissi). $\endgroup$ – Anthony Bordg Sep 27 '16 at 11:39
  • $\begingroup$ Thanks. I will try and figure out how they are the same. (To me, they are not, because I don't know much category theory and haven't studied homology theories in general, or R-modules.) $\endgroup$ – James Sep 27 '16 at 12:08
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Since you ask for a treatment of both simplicial and singular homology, I assume you have nothing against an even more general treatment that proves the required result in the general setting of $Ch(R-mod)$, the category of chain complexes with values in $R$-modules, where homology can be defined.

Both simplicial and singular homologies are particular instances of this general setting (homologies with coefficients in $R$).

Moreover, I assume you know what a category and a functor between categories are. I assume nothing beyond this. If it does not fit your background, being the tags of your question you will benefit from learning the basics of category theory.

Below I treat the case of an homotopy equivalence (deformation retract is a special case) and give the main steps (without complete proofs but with enough details).

Step 1. $H_n$ is a functor.

As you may know, being $C,\, D$ two chain complexes and $f:C\to D$ a chain morphism, $f$ induces a morphism of groups $H_n(f):H_n(C)\to H_n(D)$. One can check that $H_n$ is a functor. Actually it is useful to note that $H_n$ is an additive functor between the categories $Ch(R-mod)$ and $Ab$, meaning that the induced morphism $hom_{Ch(R-mod)}(C,D)\to hom_{Ab}(H_n(C),H_n(D))$ is a morphism of groups, i.e $H_n(f+g)=H_n(f)+H_n(g)$.

Interlude of definitions.

Def.1 (null homotopic) One says that a chain morphism $f:C\to D$ is null homotopic if there exist morphisms $s_n: C_n\to D_{n+1}$ such that $f_n = s_{n-1}\circ d_n + d_{n+1}\circ s_n$ where $+$ denotes the group law on $hom_{R-mod}(C_n,D_n)$ since the category $R-mod$ is abelian. The morphisms ${s_n}$ are called a chain contraction of $f$.

Def.2 (chain homotopic) Two morphisms $f,g$ from $C$ to $D$ are chain homotopic if $f-g$ is null homotopic.

Def.3 (chain homotopy equivalence) A morphism $f:C\to D$ is a chain homotopy equivalence if there exists $g:D\to C$ such that $g\circ f$ (resp. $f\circ g$) is chain homotopic to $id_C$ (resp. $id_D$). When $g\circ f = id_C$ one says that $f$ is a deformation retract.

Step.2 (lemma) If $f,g:C\to D$ are chain homotopic then they induce the same morphism from $H_n(C)$ to $H_n(D)$.
Proof. Since $H_n$ is additive, one has $H_n(f-g)=H_n(f)-H_n(g)$, so it suffices to prove that if a chain morphism $f:C\to D$ is null homotopic then $H_n(f)$ is the zero morphism. Let $[x]\in H_n(C)$, one has $d_n(x)=0$, hence $f_n(x)=d_{n+1}(s_n(x))\in B_n(D)$ since $f=d\circ s+s\circ d$, ie $f_n(x)$ is a n-boundary of $D$. Thus $H_n(f)([x]) = [f_n(x)] = 0$, ie $H_n(f)=0$.

Step.3 By the previous lemma and the functoriality of $H_n$, if $f:C\to D$ is a chain homotopy equivalence then $H_n(f):H_n(C)\to H_n(D)$ is an isomorphism with inverse $H_n(g)$, meaning that $f$ is a quasi-isomorphism. Indeed, one has $id_{H_n(C)}= H_n(id_C)=H_n(g\circ f)= H_n(g)\circ H_n(f)$ and $id_{H_n(D)}= H_n(id_D)=H_n(f\circ g)= H_n(f)\circ H_n(g)$.

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  • $\begingroup$ You're still left with the hard part of the proof, i.e. proving that if $f$ is a deformation retract then it induces a chain map homotopic to the identity... $\endgroup$ – Najib Idrissi Sep 22 '16 at 19:44
  • $\begingroup$ I filled in some details in case it could help James further. $\endgroup$ – Anthony Bordg Sep 23 '16 at 10:09
  • $\begingroup$ You are still missing the most important part: that if $f \simeq g$ are homotopic continuous map, then the induced maps on the singular chains are homotopic chain maps. This is the crux of the proof. $\endgroup$ – Najib Idrissi Sep 23 '16 at 11:24

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