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.

  • $\begingroup$ +1 for a well-written question. $\endgroup$ – Neal Aug 7 '19 at 16:02
  • $\begingroup$ Thank you Neal. $\endgroup$ – 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 $f\circ H\circ (f\times id)^{-1}$.

  • $\begingroup$ When I substitute to $0$ or $1$ I get $f(id_X(y))$ and $f(x_o(y))$, respectively $\endgroup$ – veronika Rmz. Aug 7 '19 at 16:17
  • $\begingroup$ Recheck! This cannot be true since $id_X(y)$ and $x_0(y)$ are undefined for $y\in Y$. Use $(f\times id)^{−1}=f^{−1}\times id$. $\endgroup$ – lulu Aug 7 '19 at 16:36
  • $\begingroup$ I do not get it $\endgroup$ – veronika Rmz. Aug 7 '19 at 16:39
  • $\begingroup$ 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 $y_0:=f\circ x_0\circ f^{-1}$. $\endgroup$ – lulu Aug 7 '19 at 17:15

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.


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