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

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.

• +1 for a well-written question. – Neal Aug 7 '19 at 16:02
• Thank you Neal. – veronika Rmz. Aug 7 '19 at 16:19

$$f\circ H:Y\times I\to Y$$ 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 $$f\circ H\circ (f\times id)^{-1}$$.
• When I substitute to $0$ or $1$ I get $f(id_X(y))$ and $f(x_o(y))$, respectively – veronika Rmz. Aug 7 '19 at 16:17
• Recheck! This cannot be true since $id_X(y)$ and $x_0(y)$ are undefined for $y\in Y$. Use $(f\times id)^{−1}=f^{−1}\times id$. – lulu Aug 7 '19 at 16:36
• Explicitely your new homotopy $G$ is given by $G(y,t)=f(H((f^{-1}(y),t)))$. This is a homotopy between $id_Y$ and the constant map $y_0:=f\circ x_0\circ f^{-1}$. – lulu Aug 7 '19 at 17:15