Discrete spaces homotopy equivalent

Question

Exercise 6.3 from Crossley 2005 Essential Topology https://link.springer.com/book/10.1007/1-84628-194-6

Prove that a discrete space consisting of $m$ points is homotopy equivalent to a discrete space consisting of $n$ points if, and only if, $m = n$.

I think the proof generalises the proposition that a 2 point space is not contractible (Proposition 6.19 in Crossley 2005), where essentially we argue that if it were contractible, then we can use the homotopy to construct a continuous map that is surjective onto the 2-point space, which is a contradiction. I've tried to formulate an inductive argument, but have failed.

Help appreciated, particularly on how to get started. I'm a beginner at this subject.

Also looking for a more elementary proof than the suggested duplicate, which has been provided below.

@VincentBoelens - No, I don't. Crossley hasn't covered it by chapter 6. He's just outlined what a homotopy is and defined homotopy equivalence and given a few examples. — Sharanjit, Jan 28 '24 at 11:50
@AnneBauval - I can see how that might be related, but I'm looking for a more elementary proof. — Sharanjit, Jan 28 '24 at 11:57
Do you know the notion of connectedness, and the fact that if a space $X$ is connected and a map $f:X\to Y$ is continuous, then $f(X)$ is connected? — Anne Bauval, Jan 28 '24 at 12:54

score 2 · Accepted Answer · answered Jan 28 '24 at 13:08

For every $n\in\Bbb N$, endow $[n]:=\{0,1,\dots,n-1\}$ with the discrete topology.

Let now $f:[m]\to[n],g:[n]\to[m]$ be two maps such that $f\circ g$ is homotopic to ${\rm id}_{[n]}$, i.e. there exists a continuous map $h:[n]\times[0,1]\to[n]$ such that $$\forall i\in[n],\quad h(i,0)=f(g(i))\text{ and }h(i,1)=i.$$ For a fixed $i$, the map $[0,1]\to[n],\;t\mapsto h(i,t)$ is continuous hence (by connectedness of $[0,1]$) its range is a connected subset of $[n]$. Moreover, it contains $h(i,1)=i$. But $\{i\}$ is the only such subset of the discrete space $[n]$. Therefore, $$\forall i\in[n],\quad f(g(i))=h(i,0)=i,$$ i.e. $f\circ g$ is not only homotopic to ${\rm id}_{[n]}$ but actually equal to it, hence $g$ is injective, so that $n\le m$ (one could alternatively argue that $f$ is surjective).

As a consequence, if $[m]$ and $[n]$ are homotopy equivalent then $m=n$.

Discrete spaces homotopy equivalent

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