The statement "If a simple graph $G$ is cubic and has a cut edge, then $\chi'(G) =4$" has a couple of proofs on this site, namely here and here. However, I was interested in a specific way of proving the statement.
In Bondy's book he leaves this problem as an exercise, but lays out a way in which to approach the exercise:
17.1.4 Deduce from Exercise 16.1.7 that a cubic graph with a cut edge is not $3$-edge-colourable.
And the mentioned exercise says the following:
16.1.7 Let $M$ be a perfect matching in a graph $G$ and $S$ a subset of $V$. Show that $\lvert M \cap \partial(S)\rvert \equiv \lvert S \rvert \pmod{2}$.
I've thought about how to apply the exercise, but I haven't found the connection Bondy laid out. I have a feeling that it might have something to do with the following: By handshake if $G$ is cubic ($3$-regular) then $2\lvert E(G)\rvert = 3 \lvert V(G)\rvert$, and since $2 \not\equiv 3 \pmod{2}$ you might get something. Another possible observation is that if the cut edge is $a\in E(G)$ then $\lvert E(G)\rvert\not\equiv \lvert E(G)\setminus\{a\}\rvert \pmod{2}$ since they would differ by $1$. But I don't see how these ideas connect to the exercise.
Can anyone give some clarification as to how one can prove the statement following Bondy's suggestion? Namely, what subset of a cubic graph should you apply $16.1.7$ to? Any and all help is greatly appreciated. Thanks for reading!
The connection to perfect matchings is just this: in a $k$-edge-coloring of a $k$-regular graph (and in particular, in a $3$-edge-coloring of a cubic graph), every color class is a perfect matching. We will take each of those color classes to be $M$, in turn.
To make use of 16.1.7, we want to choose a set $S$ with a very simple boundary $\partial(S)$. If our graph $G$ has a cut edge $e$, we can do this by letting $S$ be one of the connected components of $G-e$. If we do, then $\partial(S)$ is just the singleton set $\{e\}$: $e$ is the only edge between $S$ and $V(G) \setminus S$.
So what does 16.1.7 tell us about this scenario? It tells us that for every perfect matching $M$, we have $$|M \cap \{e\}| \equiv |S| \pmod 2.$$ Actually, $|M \cap \{e\}|$ is either $0$ or $1$, so we can be more precise and say:
Actually, in the case of cubic graphs with a cut edge, we can be certain that $|S|$ is odd, but that is irrelevant, so I will not prove it. (Exercise!) It's enough to know that all perfect matchings treat edge $e$ the same way: either all of them contain $e$, or else none of them contain $e$.
Now we can go back to our view of color classes as perfect matchings. The problem is: if we have $3$ perfect matchings that together make up a $3$-edge-coloring of $G$, then each edge is contained in exactly one of them. In particular, edge $e$ is contained in one of our perfect matchings, and not in the other two. This contradicts the claim at the end of the previous paragraph: it requires different perfect matchings to do different things with edge $e$! Therefore our $3$-edge-coloring cannot exist.
(If $\chi'(G)=4$, we do not run into this problem, because the color classes can get away with being smaller. Each color class is still a matching, but each vertex only needs to meet $3$ of the $4$ color classes, so none of the color classes need to be perfect matchings.)