# Hatcher chapter 0 exercise.

Show that $f:X \rightarrow Y$ is a homotopy equivalence if there exist maps $g,h:Y \rightarrow X$ such that $fg \simeq \mathbb{1}$ and $hf \simeq \mathbb{1}$.

Why isn't this trivial. Surely if f is a homotopy equivalence we get the maps for free with say g=h.

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 You are assuming you have these maps, not that you have a homotopy equivalence. The problem then is that you want to conclude that you can take $h=g$, which is not a priori obvious. – Alex Youcis Mar 4 '12 at 19:46 To add on Alex's comment, the implication is IF exists $h,g$ etc. THEN $f$ is homotopy equivalence. What you are asking is the other way around, you assume that $f$ is homotopy equivalence and take $g=h$; you need to prove the other direction. – Asaf Karagila Mar 4 '12 at 19:49 @AsafKaragila It still seems really trivial. – danielr1234 Mar 4 '12 at 19:52 So you have g,h and identity. Then surely, by definition f is a homotopy. – danielr1234 Mar 4 '12 at 19:53 No one said that Chapter 0 exercises are difficult, most times these exercises are meant to let you play with the definition a bit and try some easy things so you can slowly wade into the material later on. – Asaf Karagila Mar 4 '12 at 19:53

If $f_1,g_1\colon X\to Y$ are homotopic, and $f_2,g_2\colon Y\to Z$ are homotopic, then the compositions $f_2\circ f_1$ and $g_2\circ g_1$ are also homotopic.
Assuming this theorem, we show that if $fg\backsimeq1$ and $hf\backsimeq1$ then $g\backsimeq h$, so $1\backsimeq fg\backsimeq fh$ and $f$ is by definition a homotopy equivalence (with inverse homotopy equivalence $h$). To do this, we use the above theorem and the associativity of composition to find:
$$h=h\circ 1\backsimeq h(fg)=(hf)g\backsimeq 1\circ g=g$$
So $g\backsimeq h$ and we have the result as above.
 How do you prove the thing you are using. Hatcher asks to prove something like that earlier. – danielr1234 Mar 4 '12 at 20:21 Reasonable question - does Hatcher not prove it himself? It's a fairly important theorem for a lot of the rest of the book (assuming you're talking about Algebraic Topology and not some other Hatcher book). If I recall correctly, you can prove the result I mentioned directly, by taking a homotopy $H_1$ from $f_1$ to $g_1$, and a homotopy $H_2$ from $f_2$ to $g_2$, and "composing them" to get the homotopy you want. The inverted commas are because you'll need to fiddle with $H_1$ a bit (i.e. make it map to $Y\times I$ instead of just $Y$) so that $H_2\circ H_1$ is defined. – Matt Pressland Mar 4 '12 at 20:37