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?
Best Answer
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).