I am using “Matroids: A Geometric Introduction” by Gordon and McNulty.
On page 337, in Theorem 9.19 Tutte-Grothendieck invariants are defined. The theorem itself gives their description in terms of Tutte polynomial. I cannot understand how the first point of the definition (invariance) is used in the proof. I get the feeling that it is not used at all. Is it true? If so, then the first point of the definition still holds, but now this is an elementary consequence of this theorem and invariance of Tutte polynomial.
This definition is combined with the statement of the theorem, so the authors could add the first point to the definition in order to emphasize that Tutte-Grothendieck invariant is indeed an invariant, but it turned out to be messy, because it makes the impression that without this condition the theorem is false.
Thank you.
Well, the point is that conditions (3) and (4) don't even really make sense without (1): what do $f(I)$ and $f(L)$ mean? There are lots of different choices of an isthmus $I$ or a loop $L$ (which are isomorphic but not equal), and so we can't talk about a well-defined value of $f$ on them unless we know that $f$ always takes the same value on these isomorphic matroids. Similarly, in the statement of the conclusion $$f(M)=t(M;f(I),f(L)),$$ you need to know what $f(I)$ and $f(L)$ mean. If you just pick some specific $I$ and $L$ in order to interpret $f(I)$ and $f(L)$ in this conclusion, then the base case of the induction uses assumption (1): you need to know that if $M$ is an isthmus then $f(M)=f(I)$, and if $M$ is a loop then $f(M)=f(L)$.
(Separately, let me remark that there appear to be a large number of minor errors in the statement of Theorem 9.19. For instance, in conditions (3) and (4), presumably it should be assumed that $e$ is an isthmus or a loop, respectively. The domain of $f$ should consist of matroids, not isomorphism classes of matroids (if the domain is isomorphism classes, then (1) is automatic!). There should also be an assumption about the value $f$ takes on the empty matroid, which should really be the base case of the proof instead of the case $|E|=1$.)