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I've just learned about homotopy equivalence, and I'm trying to understand why, in this proof, 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
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  • Did you try leaving a comment there and waiting for a reply? – Michael Albanese Nov 19 '22 at 18:41
  • @MichaelAlbanese I've tried that in the past on several other posts and never got a reply, but I will add it as a comment there, just in case this time will be different. – Anon Nov 19 '22 at 18:43
  • Given that the person you left a comment for hasn't been active in over 3 years, I will try to address your concerns there. – Michael Albanese Nov 19 '22 at 18:46
  • @MichaelAlbanese Thanks - seeing as Hatcher hasn't introduced functors leading up to the page this claim has come from, then I still cannot see it. According to Hatcher, p. 111, the bijection should follow from the two facts: $(fg)*=fg_$ and $\mathbb{1}_*=\mathbb{1}$, and of course $fg \simeq \mathbb{1}$ and $gf \simeq \mathbb{1}$ but I don't see how to obtain the equality. If the last two had "$=$" it would be immediate. Oh - never mind, I can see my silly mistake... :/ apologies! Will close this now. – Anon Nov 19 '22 at 19:03

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I think you misread the post. What is stated is that if $f : X \rightarrow Y$ and $g : Y \rightarrow X$ are homotopy-equivalent, i.e. that $f \circ g \simeq id_Y$ and $f \circ g \simeq id_X$ then $f_* \circ g_* = id_{H_n(Y)}$ and $g_* \circ h_* = id_{H_n(X)}$. This naturally comes from the fact that $H_n$ is a functor.

Paul Cottalorda
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  • Thanks, I'm just not familiar with functors yet. See my comment above to Michael. – Anon Nov 19 '22 at 19:05
  • Which page are you on in the Hatcher? – Paul Cottalorda Nov 19 '22 at 19:08
  • p. 111, but I spotted my misunderstanding there now. I will accept your answer since it helped me realise I was looking at $id_X$ and $id_Y$ and not $id_{H_n(Y)}$ and $id_{H_n(X)}$. – Anon Nov 19 '22 at 19:11
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    Ok, on page 111, just after Proposition 2.9, Hatcher provides two properties for the induced map. (i) $(fg)* = fg$ and (ii) $\mathbb{1}* = \mathbb{1}$. This is what is meant by $H_n$ is a functor. Here, you take topological spaces and map them to groups ($X$ gives $H_n(X)$), and you take continuous functions between topological spaces and map them to homomorphisms ($f$ gives $f_*$). Think of a functor as a way to map at the same time spaces to spaces and functions to functions in a way that preserves composition (i) and identity (ii). This is a concept from category theory. – Paul Cottalorda Nov 19 '22 at 19:33