What is local index when $p$ is a regular value but not in the image of $f$?

Question

My book is From Calculus to Cohomology by Ib Madsen and Jørgen Tornehave.

This is the definition of local index:

Corollary 11.10 says if $f$ isn't surjective, then $\deg(f) = 0$, I guess by empty sum is zero. So, $\deg(f)=0$, but I don't think we necessarily have that each $\text{Ind}(f;q)$ is zero. It seems we have four cases. Which is it, and why?

Case 1: $\text{Ind}(f;q)$ is meant to be for $p$ regular value and in the image of $f$. Therefore, $\text{Ind}(f;q)$ is undefined for $p$ not in the image of $f$.

Case 2: $\text{Ind}(f;q)$ can be defined for $p$ regular value but not in the image of $f$. Vacuously, $\text{Ind}(f;q) = 1$

Case 3: $\text{Ind}(f;q)$ can be defined for $p$ regular value but not in the image of $f$. The convention is that $\text{Ind}(f;q) = 0$

Case 4: Other

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Some context:

• Is the differential at a regular point, a vector space isomorphism of tangent spaces, also a diffeomorphism of tangent spaces as manifolds?

• What might be the definition of a positively oriented chart in From Calculus to Cohomology?

• Why is there a form with compact support on a connected oriented manifold with integral one but with support contained in a given open proper subset?

Answer 1

Note that formally $p$ does not occur in $\operatorname{Ind}(f;q)$, only $q$ does. So this is defined for $q$ rather, which must be a regular point. However, it is defined only if $q$ maps to a regular value, i.e., if all "siblings" of $q$ in $f^{-1}(f(q))$ are also regular points. That is, given (regular) $q$, we find $p:=f(q)$ and if this $p$ is a regular value, we define $\operatorname{Ind}(f;q)$ as either $1$ or $-1$, otherwise (i.e., if $f(q)$ is not a regular value) $\operatorname{Ind}(f;q)$ is undefined. So your cases should really not start with $p$ in focus.