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Two topological spaces $X,Y$ have the same type of Homotopy if there exists functions $f:X\to Y$ and $g:Y\to X$ such that $f\circ g = id$ and $g\circ f = id$. In this case, $f, g$ are called Homotopy equivalences, and we denote $X\approx_f Y$.

Show that if $f:X\to Y$ is an Homotopy equivalence, then $f$ induces an isomorphism $f_{*}:H_n(X)\to H_n(Y)$, for each $n\in\mathbb{N}$.

I'm having some trouble working with this material, can anyone provide me a hint or solution?

Bernard
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michaelaba
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2 Answers2

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Hint: Well $H_n$ is a functor! So apply this functor to your compositions, and see what you can conclude!

To be specific if $f \colon X \to Y$is a homotopy equivalence with homotopy inverse $g \colon Y \to X$, then $gf \simeq 1_X$ and $fg \simeq 1_Y$. So applying the functor we get, $f_*g_*$ is the identity and $g_* f_*$ is also the identity, hence $f_*$ is an isomorphism as claimed.

Chris
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  • Why are the relations $f_g_ \simeq \mathbb{1}$ and $g_f_ \simeq \mathbb{1}$ are treated as though there's an "$=$" (in which case the bijection is immediate) instead of the "$\simeq$"? It is somehow not trivial to me that having the two composition maps homotopic to the identity map yields a bijection. What am I missing? – Anon Nov 19 '22 at 18:44
  • Note that Chris doesn't state that $f_g_ \simeq 1$ and $g_f_ \simeq 1$, but rather "$f_g_$ is the identity and $g_f_$ is the identity". – Michael Albanese Nov 19 '22 at 18:46
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You need to use (1) the functoriality of homology and (2) the homotopy invariance of singular homology which says that, if $f \simeq g : X \longrightarrow Y$ are homotopic maps, then they induce the same morphisms in all homology groups, $f_* = g_* : H_*(X) \longrightarrow H_*(Y)$. Warning: you don't have to prove these results (functoriality, homotopy invariance), just use them.

Agustí Roig
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