3
$\begingroup$

Here is the question I am trying to answer:

Show that $f: X \to Y$ is a homotopy equivalence if there exist maps $g,h: Y \to X$ such that $f \circ g \simeq 1_Y$ and $h \circ f \simeq 1_X.$ More generally, show that $f$ is a homotopy equivalence if $f\circ g$ and $h \circ f$ are homotopy equivalences.

My question is:

I do not understand how is the second part a generalization of the first part of the question, could someone explain this to me please?

$\endgroup$

1 Answer 1

5
$\begingroup$

In the second part, $f\circ g$ and $h\circ f$ are self-homotopy equivalences (maps from a space to itself which are homotopy equivalences), whereas in the first part they are required to be homotopic to the identity map. It's a generalisation because a self-homotopy equivalence is not necessarily homotopic to the identity map. For example, consider the map $\phi : S^1 \to S^1$ given by $\phi(z) = \frac{1}{z}$. It is a homotopy equivalence (with homotopy inverse $\phi$), but it is not homotopic to the identity (they induce different maps on $\pi_1$).

Said another way, every map which is homotopic to the identity is a homotopy equivalence, but the converse is not true (as demonstrated above).

$\endgroup$

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.