4
$\begingroup$

I don't understand how to solve this problem, it seems disconnected from the definition of homootpy equivalence

Let $X,Y$ be spaces with the underyling set $\{a,b\}$ for both but $\tau_=\{\phi,\{a,b\}\}$ and $\tau_Y=\{\{a\},\{b\},\{a,b\},\phi\}$. By considering paths from $a$ to $b$ in each $X,Y$, show that $X,Y$ are not homotopy equivalent.

If I recall correctly, $X,Y$ are homotopy equivalent if I can find some $f:X \to Y$ and $g:Y \to X$ such that there are homotopies $h:fg \cong id_Y$ and $k:gf \cong id_X$.

The problem is, the questions say paths within $X,Y$. Homotopy equivalence involves maps from and to $X$ and $Y$. I don't see how paths come into play at all.

Intuitively, it seems like $Y$ has the discrete topology and $X$ has the indiscrete topology so $Y$ is not path connected? While $X$ is? So...I don't recall any theorems that state that $X,Y$ must be both path-connected to be homotopy equivalent or so on... so I am stuck.

$\endgroup$
1
$\begingroup$

If $f:X\to Y$ and $g:Y\to X$ are homotopy inverses, then there is a path from $g(f(x))$ to $x$ for each $x\in X$, given by restricting the homotopy $k$ to $\{x\}\times [0,1]$. Since there is no non-constant path in $X$, this means $g(f(x))=x$ for all $x$. This severely restricts what $f$ and $g$ can be, and you should be able to finish from here (note that not very many maps $f:X\to Y$ are continuous).

Much more generally, suppose $X$ is any topological space and let $\pi_0(X)$ denote the set of path-components of $X$. Note that any continuous map $f:X\to Y$ induces a map $f_*:\pi_0(X)\to\pi_0(Y)$ by sending a path-component of $X$ to the path-component of $Y$ containing its image under $f$. Note also that $(gf)_*=g_*f_*$ for maps $f:X\to Y$ and $g:Y\to Z$ and that if $f:X\to X$ is the identity map, then $f_*:\pi_0(X)\to\pi_0(X)$ is also the identity map. Furthermore, note that if $f$ is homotopic to $g$, then the maps $f_*,g_*:\pi_0(X)\to \pi_0(Y)$ are equal, since there is a path in $Y$ from $f(x)$ to $g(x)$ for each $x\in X$. (All of this can be summed up by saying that $\pi_0$ is a functor from the category of spaces and maps up to homotopy to the category of sets and functions.)

Now suppose $f:X\to Y$ and $g:Y\to X$ are homotopy inverses. Then since $fg\cong 1_Y$ and $gf\cong 1_X$, we have $f_*g_*=1_{\pi_0(Y)}$ and $g_*f_*=1_{\pi_0(X)}$. This means that $f_*:\pi_0(X)\to\pi_0(Y)$ and $g_*:\pi_0(Y)\to\pi_0(X)$ are inverse bijections. In particular, $X$ and $Y$ must have the same number of path-components.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.