How can I show that circle does not have the same homotopy type as any finite space?

Question

We know that circle has the same weak homotopy type as pseudocircle, which is finite.

I wonder is there a quick way to show that no finite space can have the same homotopy type as a circle?

Answer 1

If $X$ is a finite space and $f : S^1 \to X$ is a homotopy equivalence with homotopy inverse $g : X \to S^1$, then the map $g \circ f : S^1 \to S^1$ must be homotopic to the identity. But by degree theory, a non-surjective map $S^1 \to S^1$ cannot be homotopic to the identity, and $g \circ f$ clearly is not surjective.

Answer 2

More generally, if $X$ is a finite space and $x,y\in X$, define $x\leq y$ if $y\in\overline{\{x\}}$ and let $\sim$ be the equivalence relation generated by this relation $\leq$. Note that if $f:X\to Y$ is any continuous map from $X$ to a $T_1$ space $Y$, then $x\geq y$ implies $f(x)=f(y)$ and thus $f$ is constant on the equivalence classes with respect to $\sim$. These equivalence classes are closed in $X$ (since the equivalence class of $x$ contains $\overline{\{x\}}$) and thus also open since there are finitely many of them, so the quotient space $X/{\sim}$ is discrete.

The upshot, then, is that any continuous map from a finite space to a $T_1$ space factors through a discrete space. In particular, it must induce trivial maps on $H_n$ and $\pi_n$ for $n>0$. So, if $Y$ is a $T_1$ space with $H_n$ or $\pi_n$ nontrivial for some $n>0$, then $Y$ cannot be homotopy equivalent to a finite space.

(In fact, a bit more strongly, you can show that the induced map $X/{\sim}\to Y$ will also be a homotopy equivalence if $X\to Y$ is, so $Y$ must be homotopy equivalent to a discrete space.)