I have been reading notes on Feynman diagrams and homological integration here: https://math.berkeley.edu/~erabin/The%20Divergence%20Complex%20and%20Feynman%20Diagrams.pdf.
However, I'm confused about the proof of Theorem 4.1. I do not understand why it is the case that if $i(\sigma(\Phi(v)))$ is a vertex in $\Gamma$ that is identified with a vertex of $\Gamma'$ then one could pair $v$ with the half-edge corresponding to $\sigma(\Phi(v))$ (this statement is located in the last five lines of page 7). Shouldn't it be possible that $i(\sigma(\Phi(v)))$ is identified but there does not exist a half-edge which corresponds to $\sigma(\Phi(v))$?
My original guess is to first use $\Delta_m^\mu$ with $m=1$ then $\Delta^\kappa$ but turns out we only consider the operations $\Delta_m^\mu$ for $m\geq 2$.