# Stokes Theorem Manifold with Corners Proof

I'm working through the proof for Stokes' Generalized Theorem for Manifolds and have a questions about corners. I've seen several proofs for manifolds with corners by creating diffeomorphisms to spaces like $(0,1] \times (0,1] \times (0,1)^{n-2}$, but I was wondering why this is necessary.

For example, in the diagram below, it seems like I'm able to parse out this figure in the corner using two spaces which are diffeomorphic to just $(0,1] \times (0,1)$ rather than $(0,1] \times (0,1]$. Can't this always be done? Basically can't any corner be reduced to the case of something diffeomorphic to an open cube with one side of the boundary?

• If I am understanding your picture correctly, the two spaces that you are joining together do not include the common face (which happens to contain the corner). How would you account for this face? – user197427 May 4 '15 at 20:17
• That's what I was thinking too. If you don't like corners in your manifolds (which is a perfectly reasonable stance), you might want to look into "straightening corners". – HSN May 4 '15 at 20:20