# Sufficient condition for $f$ to be a homotopy equivalence.

Let $$f:X\rightarrow Y$$ be a continuous map of topological spaces. Prove that $$f$$ is homotopy equivalent if there exists continuous maps $$g,h:Y\rightarrow X$$ such that $$f\circ g\simeq Id_Y$$ and $$h\circ f\simeq Id_X$$.

My class has just started on homotopy theory so I'm sure that this problem doesn't require much more than the definitions. In light of this what the problem statement tells me is that since $$h\circ f\simeq Id_X$$ there is a continuous map $$F:X\times I\rightarrow X$$ where $$I=[0,1]$$ with \begin{align*} F(x,0)&=(h\circ f)(x)\\ F(x,1)&=Id_X(x). \end{align*} Similarily since $$f\circ g\simeq Id_Y$$ there is a continuous map $$G:Y\times I\rightarrow Y$$ with \begin{align*} G(y,0)&=(f\circ g)(y)\\ G(y,1)&=Id_Y(y). \end{align*}

Now with my current understanding if I wish to show that $$f$$ is a homotopy equivalence then I need to come up with a continuous mapping I've called it $$\alpha:Y\rightarrow X$$ such that $$f\circ \alpha\simeq Id_Y$$ and $$\alpha\circ f\simeq Id_X$$. Of course this amounts to coming up with continuous mappings $$H:X\times I\rightarrow X$$ and $$J:Y\times I\rightarrow I$$ such that \begin{align*} H(x,0)&=(\alpha\circ f)(x)\\ H(x,1)&=Id_X(x)\\ J(y,0)&=(f\circ\alpha)(y)\\ J(y,1)&=Id_Y(y). \end{align*}

Essentially what it looks like I need to do is define $$\alpha$$ using the $$g$$ and $$h$$ in such a way that all the desired mappings are continuous but so far I'm at a loss especially since none of the given mappings are invertible keeping $$\alpha$$ well defined also seems like a big concern. Any help to get going in the right direction is greatly appreciated.

• You don't need to "keep $\alpha$ well defined", you just need to choose an appropriate function $\alpha : Y \to X$. Given what you have in front of you, what other choice is there but $\alpha = g$ or $h$? So your job becomes: choose one of them (by symmetry of the setup, if one works then the other will), and prove what you need to prove. – Lee Mosher Jan 13 '20 at 16:27

## 1 Answer

I will first prove two lemmas. These are obvious results, and you should use these extensively.

Lemma 1. Let $$\alpha,\beta:X \to Y$$ and $$\gamma:Y \to Z$$ be continuous maps. If $$\alpha \simeq \beta$$ then $$\gamma \alpha \simeq \gamma \beta$$.

Proof. If $$F$$ is a homotopy between $$\alpha$$ and $$\beta$$, then $$\gamma F$$ is a homotopy between $$\gamma \alpha$$ and $$\gamma \beta$$, as you may easily check.

Lemma 2. Let $$\alpha:X \to Y$$ and $$\beta, \gamma:Y \to Z$$ be continuous maps. If $$\beta \simeq \gamma$$ then $$\beta \alpha \simeq \gamma \alpha$$.

Proof. If $$G$$ is a homotopy between $$\beta$$ and $$\gamma$$, then $$G \circ (\alpha \times id_I):X \times I \to Y \times I \to Z$$ is a homotopy between $$\beta \alpha$$ and $$\gamma \alpha$$.

Now we claim that $$g \simeq h$$. This will imply $$gf \simeq hf \simeq id_X$$, and then combining with the assumption $$fg \simeq id_Y$$, the result follows. But our claim is obvious since $$h=h\circ id_Y \simeq h(fg)=(hf)g \simeq id_X \circ g=g$$.

• Thank you. This is exactly what my intuition was trying to tell me but I couldn't quite get it on paper. I appreciate the insight. – Walt Jan 13 '20 at 20:47