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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?

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  • $\begingroup$ Do you want a continuous action? $\endgroup$ – Nitrogen Feb 10 '16 at 22:53
  • $\begingroup$ @Nitrogen Yes, a continuous action. $\endgroup$ – Cihan Feb 10 '16 at 22:55
  • $\begingroup$ What is $C_2$? Is it $\mathbb{Z}/2\mathbb{Z}$? $\endgroup$ – mad_algebraist Feb 10 '16 at 22:56
  • $\begingroup$ I am not entirely sure but a counter-example could be $S^1$ glued with two lines $(0,1)\to (2,0)$ and $(0,-1)\to (2,0)$, with the antipodal action on $S^1$ and the reflexion on the $x$-axis for the two lines. Then the only fixed point is $(-1,0)$ and I think that the quotient space is contractible. $\endgroup$ – Nitrogen Feb 10 '16 at 23:19
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    $\begingroup$ @Nitrogen Is'nt the quotient space in this case a circle glued with a segment? I mean, it's not contractible. $\endgroup$ – lisyarus Feb 11 '16 at 0:10
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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 $S^1\vee S^1$).

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