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Let $M$ be a $m$-dimensional compact manifold without boundary and $W(M)$ the non-compact $CW$-complex obtained by glueing $[0,1)\times (0,1)^{m-1}$ to $M$, identifying the boundary $\{0\}\times(0,1)^{m-1}$ with a hyperfurface in a chart of $M$ as in the following picture.

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Let $\Sigma_k$ be the symmetric group on $k$-letters and $F(M,k)$ the $k$-th configuration space consisting of ordered $k$-tuples of distinct points in $M$. Let $\Sigma_k$ act by permuting the coordinates. There exist a $\Sigma_k$-equivariant inclusion $$ i: F(M,k)\longrightarrow F(W(M),k) $$ such that the diagram commutes

enter image description here

Question: Is $i$ a homotopy equivalence? Is $$i/\Sigma_k: F(M,k)/\Sigma_k\longrightarrow F(W(M),k)/\Sigma_k$$ a homotopy equivalence?

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Let me complete @Michael's answer. The space $F(S^1,2)$ is a torus with the diagonal removed. Picture it in $\mathbb R^3$ as usual. It deformation retracts to a shifted diagonal, which is again an $S^1$.

Now add those configurations where the first point is on the line. You do this by gluing an annulus "inside" the torus passing through the point $(*,*)$, where $*$ is the gluing point, then take away $(*,*)$. If the second point enters the interval as well, you get points in a solid triangle glued to the annulus, which does not affect the homotopy type.

Now you reverse the role of both points, and glue another annulus into the space that would normally be reserved for a person in a lifebuoy, and add the extra triangle if you like. Again, this annulus passes through $(*,*)$, which you keep removed.

There is a $\Sigma_2$ equivariant deformation retraction into a circle with two circles glued in at antipodal points, so $\pi_1(F(S^1\cup I,2))\cong F_3$ is a free group with $3$ generators. If you divide by $\Sigma_2$, the quotient has $\pi_1$ equal to the free group with two generators. In both cases, the result is different from what you get for $S^1$. The reason is that while one points escapes into the interval, the second one can run freely around the circle.

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    $\begingroup$ $\pi_3$ or $\pi_1$? $\endgroup$ – Michael Jan 9 '16 at 0:00
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I don't think so, even in the case of $M=S^1$ and $k=2$: $F(M,2)$ would be homotopy equivalent to $S^1$: pick the 1st point on the circle, and the 2nd point can be only on $S^1$ \ $pt$, which makes $F(S^1,2)$ an open interval fibration over $S^1$, thus homotopy equivalent to $S^1$.

On the other hand, when you pick the 1st point on $W(S^1)$ there are 3 possibilities: you pick it on the gluing point, you pick it on the circle but not on the gluing point, or you pick it on the open interval connected to the circle via the gluing point. Wait a sec, I have doubts now: have to figure out how exactly those parts are glued together...

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