Suppose $f$ is a continuous function from a unit cube $[0,1]^n$ to itself, then $f$ has at least a fixed point. Further suppose $f$ is smooth, $0$ is a regular value of $f(x)-x$, and the fixed points are possibly on the boundary. In reading some textbook, I find that the fixed points on the boundary can not be involved in the fixed-point index. I wonder whether there is an index theorem that treats the fixed points on the boundary?
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2$\begingroup$ The Leftschetz theorem says that there is some local formula. How to actually compute the local index is the subject of a wide literature, at many levels of generality. I haven't read much of it, but I liked Goresky-MacPherson. That's probably more generality than you need, but it is very well written. $\endgroup$– Ben WielandCommented Jun 21, 2019 at 0:01
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If you consider a self-map $f:M \to M$ a compact manifold with boundary then the map can be deformed outside the boundary and there are no fixed points on the boundary.
If you consider the map (and deformations) preserving the boundary $(f,\partial f) :(M,\partial M) \to (M,\partial M)$ then you may use the fixed point index of the restriction $\partial f :\partial M \to \partial M$ to detect fixed points on the boundary.
