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. $\endgroup$ 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 $H(f)$. $\endgroup$ 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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.