I wish to prove the following result:
Let $M$ and $M'$ be disjoint matchings in a graph $G = (V,E)$. Show that every component of the graph $(V, M \cup M')$ is a path or an even circuit.
We just started learning about matchings, which is a set of edges that pairwise have no endpoint in common. Formally it means that if $M$ is a matching, we know that $M \subseteq E$ and for $e, e' \in M$ it holds that $e \cap e'= \emptyset$.
Now I need to make some sort of disctinction for union of two of these matchings $M$ and $M'$, such that $M \cap M '= \emptyset$. I do not see what kind of distinction to make, I also feel a little bit overwhelmed by the "every component" part. It feels very general. Can somebody offer a useful hint, how to proceed/start a good distinction? Basically what I think I need to do is show that if $C$ is a circuit, it has to be even.
scrap paper:
Suppose we colour matching $M$ blue and we colour matching $M'$ red. Then the idea is that for an even circuit, these two colours alternate.
Edges in $M$ do not intersect and edges in $M'$ do not intersect either. The only way edges in $(V(G),M\cup M')$ can intersect is between edges in $M$ and edges in $M'$. We can't have three edges intersect at the same vertex. That is because two of those edges both belong to either $M$ or $M'$, But they can't intersect by definition of a matching. Hence the degree of each vertex in $(V(G),M\cup M')$ is at most two. Hence vertices in each components is at most 2. So then that means the component is either a path or a cycle.