A small question about $M/\partial M$, where $M$ is a manifold. I have a question about topology.
Suppose $M$ is a space with boundary $\partial M$, is that possible to obtain a space with the quotient topology $M/\partial M$? If this is possible, does $M/\partial M$ have a boundary?
I guess this quotient space $M/\partial M$ makes sense and it has no boundary, i.e. $\partial(M/\partial M)=0$. This works for $M$ being a disk $D$, whose boundary is a circle $S$.
Hope you can help me.
 A: This is a duplicate of When does the quotient of a manifold with boundary become a manifold? , but I don't find the answers there fully satisfactory.
As mentioned earlier in the comments, the boundary $\partial M$ of a smooth manifold $M$ has a collar neighbourhood $\partial M \subset U \simeq \delta M \times [0, 1)$. This gives us a neigbourhood of our singular point in $M/\partial M$, which I will denote by $x$ $(= [\partial M] \in M/\partial M$. So, in $M/\partial M$, we have a neighbourhood $x \in \tilde{U} =  (\partial M \times [0, 1))/\partial M$. 
The local homology $H_\star(\tilde{U}, \tilde{U} - x)$ should be isomorphic to the homology of $S^n$ if $M/\partial M$ was a manifold. We have a long exact sequence of reduced homology:
$\ldots \to \tilde{H}_k(\tilde{U}) \to \tilde{H}_k(\tilde{U},\tilde{U}-x) \to \tilde{H}_{k-1}(\tilde{U}-x) \to \tilde{H}_{k-1}(\tilde{U}) \to \ldots$
It's easy to see that $\tilde{U}$ is contractible (as it's basically cone over $\partial M$), so by exactness we get isomorphisms $\tilde{H}_k(\tilde{U},\tilde{U}-x) \simeq \tilde{H}_{k-1}(\tilde{U}-x)$. As $\tilde{U}$ is a cone over $\partial M$ and $x$ is a cone point, $\tilde{U} - x$ deformation retracts onto $\partial M$, so $\tilde{H}_{*+1}(\tilde{U}, \tilde{U}-x) \simeq \tilde{H}_*(\tilde{U} - x) \simeq \tilde{H}_*(\partial M)$
As local homology of $x$ is supposed to be the same as homology of $S^n$ if $M/\partial M$ is has to be manifold, the necessary condition for $\partial M$ is that $\partial M$ is a homology sphere. I'm quite sure that it actually needs to be an actual sphere, though I don't see how to show this.
