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It is a general fact that the boundary $\partial\mathcal{M}$ of any topological manifold with boundary $\mathcal{M}$ is itself a topological manifold and has empty boundary, i.e. $\partial (\partial\mathcal{M})=0$.

Now, I am a little bit confused about the following: There is also the notion of manifolds with corners, i.e. second countable Hausdorff spaces, which are locally homeomorphic to $\mathbb{R}_{\geq 0}^{d}=[0,\infty)^{d}=\{x\in\mathbb{R}^{d}\mid x_{i}\geq 0\}$. I am aware of the fact that topologically, manifolds with corners are one and the same as topological manifolds with boundary, since the space $\mathbb{R}_{\geq 0}^{d}$ is homeomorphic to the model space of manifolds with boundary, i.e. the half-plane $\mathbb{H}^{d}=[0,\infty)\times\mathbb{R}^{d-1}$. This is no longer true in the differentiable setting, since these two sets are not diffeomorphic. However, in my question I do not care about differentiability, but I am only thinking about the topological category.

So, in other words, every manifold with corners is in particular a (topological) manifold with boundary. So, it should also be true that the boundary of every such manifold is by itself a manifold with empty boundary, right? I mean, the Theorem mentioned at the beginning is for arbitrary manifolds with boundary and the notion of boundary is a purely topological object and has nothing to do with a differentiable structure. Now, when I think for example about the solid cylinder, i.e. the manifold with corner defined by $C:=D^{2}\times [0,1]$, where $D^{2}$ is the disk (=closed 2-ball), then its boundary is clearly

$$\partial C=(\partial(D^{2})\times [0,1])\cup (D^{2}\times\partial [0,1])=(S^{1}\times [0,1])\cup (D^{2}\times\{0,1\}).$$

But this is not a topological manifold without boundary right? (Or maybe it is, and I just have an thinking error. But then it has to be homeomorphic to one of the compact surfaces, by the classification theorem and since it has no holes, it has to be homeomorphic to $S^{2}$, but I can't see why this should be the case. I already tried to choose triangulations of the boundary cylinder $\partial C$, but I do not find a Euler-characteristic of $2$. For example, take a single prism as a cellular decomposition of the solid cylinder. Then its boundary has $5$ faces, $7$ edges and $6$ vertices and hence, I would get $\chi=4$).

EDIT: I know that there is a related question here on this side, however, this question seems to ask about the differentiable category and does not answer my question and confusion about the solid cylinder.

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Well, manifolds with corners are not one and the same as topological manifolds. Let me come back to that point later, and address the easy parts of your question first.

In fact $\partial C=(S^{1}\times [0,1])\cup (D^{2}\times\{0,1\})$ is indeed homeomorphic to $S^2$. One way to see this is to subdivide $S^2$ at two latitude lines: the arctic circle at $67^\circ$ north latitude and antarctic circle at $67^\circ$ south latitude. The required homeomorphism takes the northern polar cap above $67^\circ$ north latitude to $D^2 \times \{1\}$; it takes the southern polar cap to $D^2 \times \{0\}$; and it takes everything between the arctic and antarctic circles to $S^1 \times [0,1]$. You can write down a completely concrete formula for this homeomorphism using a piecewise smooth formula with three pieces, expressed in terms of spherical coordinates on $S^2$ and polar coordinates on $D^2$.

However, this does not give a manifold-with-corners structure to $S^2$: the "corner structure" might appear to be the arctic and antarctic circles; but those are circles, they are disjoint from the topological boundary (which is actually empty), and the only "corners" on a 2-manifold-with-corners are isolated points on the topological boundary.

In fact the only manifold-with-corners structure on $S^2$ is one whose charts have values in $(0,\infty)^2$, which is equivalent to an actual topological manifold structure (with empty boundary).

As for your prism structure, you simply miscounted the edges, there are 9 edges: three around the arctic circle; three cutting across from the arctic circle to the antarctic circle; and three around the antarctic circle.

So what about your one and the same comment? It helps here to think in terms of category theory. Intuitively, a manifold with corners has more structure than a topological manifold. By forgetting this structure, you get a forgetful functor from the category of manifolds with corners to the category of topological manifolds.

What exactly has been forgotten? You write that the model space $[0,\infty)^d$ for a manifold with corners is homeomorphic to the model space $[0,\infty) \times \mathbb R^{d-1}$. This is true. However, the definition of manifold with corners has a rather strong restriction on overlap maps, which you have not mentioned. If the $U,V \subset [0,\infty)^d$ are the open subsets that are the targets of two charts in your manifold-with-corners atlas, and if $\psi : U \to V$ is the overlap map itself, then not only must $\psi$ be a homeomorphism (from $U$ to $V$), but $\psi$ must respect the corner structures. I suggest that you read exactly what atlases are, focussing on the overlap condition, in whatever textbook you might have on this topic. But in brief: the map $\psi$ must take the origin in $U$ to the origin in $V$ (assuming $U$ or $V$ contains the origin); it must take the union of the coordinate axes (intersected with $U$) to the union of the coordinate axes (intersected with $V$); it must take the union of the coordinate 2-planes to the union of the coordinate 2-planes; and so on. All of this extra structure is forgotten when you pass to the category of topological manifolds.

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  • $\begingroup$ So in summary, for manifolds with corners in general $\partial (\partial M)$ need not be empty but would rather be the collection of corners. $\endgroup$
    – quarague
    Jan 18 at 14:17
  • $\begingroup$ Frankly, I am unsure of how to define $\partial(\partial M)$ when $M$ is a manifold-with-corners. One is somehow forced out of the category when trying to formulate the boundary of a manifold-with-corners. Or perhaps instead one has to rip apart the topological boundary in order to form the manifold-with-corners boundary. $\endgroup$
    – Lee Mosher
    Jan 18 at 14:19
  • $\begingroup$ Don't you get a natural boundary definition where the boundary of a manifold-with-corners is also a manifold-with-corners? This should be an extension of smooth manifolds, so all the maps are smooth whereever it makes sense for them to be smooth. $\endgroup$
    – quarague
    Jan 18 at 14:22
  • $\begingroup$ One probably needs still more general categories in order to have a good boundary theory, such as the category of stratified spaces, but that's way outside the bounds of this post. $\endgroup$
    – Lee Mosher
    Jan 18 at 14:23
  • $\begingroup$ Thats a great answer! Thank you very much, this clears it up for me :-) And yes indeed, I have embarssingly miscounted the number of edges. I should probably draw a picture in the future^^. $\endgroup$
    – B.Hueber
    Jan 18 at 15:22

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