Is there a version of the Poincaré–Hopf theorem for manifold with corners? As we know, the square $S=[0,1]\times[0,1]$ is not a manifold with boundary. Instead, it's a manifold with corners. For a tangent vector field on a compact manifold with boundary, we have the Poincaré–Hopf index theorem.
Then here is my question: is there a similar index theorem for manifolds with corners (e.g., the square, the cube) which relates the indices of a tangent vector field to the Euler characteristics of the manifold? If affirmative, what is the statement of this theorem?
Btw, I find an economics textbook (Vives, X., 1999. Oligopoly pricing: old ideas and new tools. MIT Press) which states that the Poincaré–Hopf theorem can apply to a compact cube (a manifold with corners). The statement (on page 362) is as below:

This "version" is frequently used in economics papers. Is it correct?
 A: With regards to the updated question: Note that the quoted statement is that the vector field points inward at the boundary. In particular this means that there are no singularities at the corners (nor at the flat parts of the boundary). Thus you can simply replace the manifold by a slightly smaller one, for which the corners are "rounded off". This then is a manifold with boundary, to which the usual Poincaré-Hopf theorem applies.
That being said, as discussed in the comments, the situation gets a bit more interesting when there are singularities at the boundary. There, the theorem still holds, provided the notion of index is correctly defined for those singularities. I think for this, one can simply copy the usual proof of the Poincaré-Hopf theorem which cuts out small balls around the singularities and then simply define the index as whatever is needed to make this work.
Toying around with the Gauss-map, I would expect something like
$$ \operatorname{ind}(x_0) = \lim_{\epsilon \to 0} \frac{1}{|S^{n-1}|} \int_{S\cap \partial B_\epsilon(x_0)} \left(\frac{u}{|u|}\right)^* \omega_{S^{n-1}}$$
but for corners there might be situations where this limit is not entirely well defined.
