If $X\approx Y$ and $X$ is contractible then $Y$ is contractible

Question

If $X\approx Y$ and $X$ is contractible then $Y$ is contractible.

Attempt to the solution

Since $X\approx Y$, then there exist a continuous and bijective function $f:X\to Y$ such that $f^{-1}:Y\to X$ is continuous.

Since $X$ is contractible then $id_X\simeq x_o,$ where $x_o$ is a constant function, that is there exist an homotopy $H:I\times I\to X$ such that $H(x,0)=id_X(x)$ and $H(x,1)=x_o(x).$

Now to find $id_Y\simeq y_o$ I did try to draw a diagram

$\require{AMScd}$ \begin{CD} X\times I @>f\times id>> Y\times I\\ @V H V V @VV f\circ H V\\ X @>>f> Y \end{CD}

I am not sure whether if its correct or not.

From here how could I define the homotopy composition $f\circ H,$ $id_Y\simeq y_o$ ?

If someone could help me, thank you.

Answer 1

$f\circ H:Y\times I\to Y$ makes no sense since $H$ is defined on $X\times I$.

Instead notice that $f\times id$ is a homeomorphism so you can let your homotopy be $f\circ H\circ (f\times id)^{-1}$.

Answer 2

If you have proved that homotopy equivalence is an equivalence relation, all that is needed is to show that being contractible is exactly the same as being homotopy equivalent to a point.