Does $X^{C_2} \simeq * \simeq X/{C_2}$ imply $X \simeq *$?

Question

What the title says. Let $C_2$ be the cyclic group of order 2, and $X$ be a topological space with a $C_2$-action (acting continuously) such that both the quotient space $X/{C_2}$ and the subspace of fixed points $X^{C_2}$ are contractible. Does that force $X$ itself to be contractible?

Answer 1

Here is a counterexample, though it may not be very interesting to you because the space involved is not Hausdorff. Let $X=\{a,b,c,d,e\}$, with topology generated by the sets $\{a\}$, $\{b\}$, $\{c\}$, $\{d,a,b,c\}$, and $\{e,a,b,c\}$. Define $\sigma:X\to X$ by $\sigma(a)=b$, $\sigma(b)=a$, $\sigma(c)=c$, $\sigma(d)=e$, and $\sigma(e)=d$. Then $\sigma$ is a homeomorphism and gives an action of $C_2$ on $X$. The fixed points and quotient are contractible (the fixed points are just $\{c\}$ and the quotient has a point $[d]$ that is in the closure of every point), but $X$ is not contractible (in fact, it has the weak homotopy type of $S^1\vee S^1$).