I have to prove that if X an Y are contractibles, then any continuous map between them is an homotopic equivalence.
I have written the following:
"As $X$ and $Y$ are contractibles, we can consider the map $f$ between $x\in X$ and $y\in Y$, both fixed. We are going to show that $x\simeq y$. Let be $f$ the map that to $x$ associates $y$: is continuous, because the domain is discrete. Similarly let be $g$ the map that to $y$ associates $x$, is continuous. Also, $f\circ g\simeq Id_{\{y\}}$ and $g\circ f\simeq Id_{\{x\}}$. So, $f$ is an homotopic equivalence."
Well, really I am not sure at all, because I know that this is not the condition of homotopic equivalence, but, I don't find any way to do it. I feel that I am close, but I don't know how to finish.
Sorry if I've commited one error.
Thanks for the answers.
You are on the right tracks with your choice of $x$ and $y$. Let's denote those by $x_0, y_0$. If $f\colon X\to Y$ is any map and $g\colon Y\to X$ is the constant map taking every point of $Y$ to $x_0$, then you can define the composites $g\circ f\colon X\to X$ and $f\circ g\colon Y\to Y$. Now, since $X$ is contractible, we may assume there is a map $H\colon X\times I\to X$ such that $H(x, 0)=x, H(x, 1)=x_0$ for all $x\in X$ (otherwise we just redefine $g$). Now, notice that $H(x, 1)=(g\circ f)(x)$, so $H$ is a homotopy from the identity of $X$ to $g\circ f$. A similar homotopy between the identity of $Y$ and $f\circ g$ is pretty much the same, but now $(y, 0)\mapsto y, (y, 1)\mapsto y_0$.
You were very close :)