4
$\begingroup$

Consider the space $X=\{0\}\cup \{\frac{1}{n},n\in \mathbb{N}\}$ with the topology induced by the real line.

Is $X$ homotopy equivalent to some enumerable discrete space $Y$?

My try was the following: Let $Y$ a enumerable discrete space. If $X$ and $Y$ are homotopy equivalent then there exists maps $f:X\to Y$ and $g:Y\to X$ s.t $g\circ f:Y\to Y$ is homotopic to $Id_Y$. But $X$ is compact, so the image of $g\circ f$ must be finite.

However I can't explore this ideia :(

$\endgroup$
  • $\begingroup$ Hint: 1. Show that if $Y$ is discrete and $f:Y\rightarrow Y$ is any function, then any homotopy of $F$ is constant in time. 2. As you noticed, if $f:X\rightarrow Y$ is continuous, then $f(X)$ is a finite subset of $Y$. Now, if $g:Y\rightarrow X$ then $f\circ g$ has finite image. Now apply 1... ( Feel free to write up your own answer. And note that this proof works for any infinite $Y$, but does not work if $Y$ is finite.) $\endgroup$ – Jason DeVito Feb 2 '16 at 2:34
  • $\begingroup$ @JasonDeVito Than you, nice solution, I was intended to fill you sketch, but Najib Idrissi came first. $\endgroup$ – O Empalador de Cabras Feb 2 '16 at 19:27
3
$\begingroup$

Suppose that you have a homotopy equivalence $f : X \to Y$ with homotopy inverse $g : Y \to X$, where $Y$ is some discrete space (of any cardinality, finite or infinite). The space $X$ is compact, thus $f(X) \subset Y$ is compact; but the only compact subspaces of a discrete space are finite, so $f(X)$ is finite.

Since $f$ and $g$ are inverse homotopy equivalences, $f \circ g$ is homotopic to $\operatorname{id}_Y$, i.e. there is a homotopy $H : Y \times [0,1] \to Y$ with $H(y,0) = f(g(y))$ and $H(y,1) = y$. But $Y$ is discrete, so $H$ has to be constant on connected subspaces, and $\{y\} \times [0,1]$ is connected, thus $f(g(y)) = y$. But the image of $f$ is finite, whereas the image of the identity is all of $Y$, so $Y$ is finite.

But $X$ has an infinite number of path-connected components (every singleton is a path-connected component), whereas a finite discrete space only has a finite number of path-connected components. Since a homotopy equivalence preserves the number of path-connected components, this is a contradiction.

$\endgroup$

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.