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

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?

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.
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.)