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Let $X$ be a topological space. For $n \ge 2$, the group $S_n$ acts naturally on $X^n$ as $\sigma . (x_1,...,x_n)=(\sigma(x_1),...,\sigma (x_n)), \forall (x_1,...,x_n) \in X^n$ . So we can consider the quotient space $X^n /S_n$.

Now let $X$ be a Hausdorff, contractible, locally path connected topological space. Then is the quotient space $X^2/S_2$ also contractible ? If it is not in general, then is the fundamental group at least trivial ? ($X^2/S_2$ is obviously path connected ) What about $X^n/S_n$ in general ?

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The space you're looking at is called the "symmetric product", $\operatorname{SP}^n(X) = X^n / S_n$.

It is a functor and preserves homotopies, so yes, the $n$th symmetric product of a contractible space is contractible, for all $n$ and with no assumptions on the space. The proof is rather immediate. Suppose that $H : X \times [0,1] \to X$ is a homotopy such that $H(x,0) = x$ and $H(x,1) = y$ for a fixed $y_0 \in X$. Define a new homotopy: $$\begin{align} H_n : X^n/S_n \times [0,1] & \to X^n/S_n, \\ [x_1, \dots, x_n] & \mapsto [H(x_1,t), \dots, H(x_n, t)]. \end{align}$$ You need to check that $H_n$ is well-defined and continuous, which is immediate. Then it's clear that, for $[\mathbf{x}] \in X^n/S_n$, you have $H_n([\mathbf{x}], 0) = [\mathbf{x}]$ and $H_n([\mathbf{x}],1) = [y_0,\dots,y_0]$. So $H_n$ is a homotopy between the identity of $X^n/S_n$ and a constant map. Hence $X^n/S_n$ is contractible.

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  • $\begingroup$ Could you please elaborate on why $H_n$ is continuous ? $\endgroup$ – user Apr 14 '18 at 10:25
  • $\begingroup$ @users By the definition of the quotient topology. $\endgroup$ – Najib Idrissi Apr 14 '18 at 13:08

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