This is stated as a corollary from the Homotopy Axiom (Rotman) which states the if $f \sim g$ are homotopic maps, then $H_n(f) = H_n(g)$ for $n \geq 0$. (going to drop the $n$)

Because $X$ and $Y$ have the same homotopy type, there exist $f: X \to Y$ and $g: Y \to X$ such that $f \circ g \sim id_Y$ and $g \circ f \sim id_X$. By the Axiom, $$id_{H(Y)} = H(id_Y) = H(f\circ g)$$ $$H(g \circ f) = H(id_X) = id_{H(X)}.$$

So $H(f) \circ H(g) = id_{H(Y)}$ and $H(g) \circ H(f) = id_{H(X)}$

Aren't these just "left" and "right" inverses? How do we draw the isomorphism?

  • 2
    $\begingroup$ You mean $f\circ g$ is homotopic to $\text{id}_Y$ etc. $\endgroup$ – Lord Shark the Unknown Jul 13 '18 at 6:38
  • $\begingroup$ @LordSharktheUnknown sort yes, that's what i meant a typo $\endgroup$ – Hawk Jul 13 '18 at 6:53
  • $\begingroup$ So you have obtained that $H(f)$ is an isomorphism between $H(X)$ and $H(Y)$. $\endgroup$ – C.Ding Jul 13 '18 at 7:22
  • $\begingroup$ @C.Ding not quite. It doesn't have a real inverse. $\endgroup$ – Hawk Jul 13 '18 at 7:27
  • $\begingroup$ @Hawk: consider two groups $G$ and $H$ and a homomorphism $f \colon G \to H$. What does it mean for $f$ to be an isomorphism? Now replace $G$ by $H(X)$, $H$ by $H(Y)$ and $f$ by $H(f)$. $\endgroup$ – Magdiragdag Jul 13 '18 at 7:33

There is nothing left to be shown. You have a group homomorphism $H(f)$ which has an inverse, namely $H(g)$, and that is the definition of an isomorphism (in any category).


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