Maximal matching for a Biparitie Graph is the maximum cardinality set of edges such that no two edges share any vertex.
We are given a biparitie graph and let's say there is only one edge $E$ that has vertex $X$ as one of its end-point.
I feel it is always safe to include the edge $E$ in the maximal matching because the given graph is biparitite in nature.
Can we prove this? I have worked on the proof which seems to justify this, but I am not satisfied with it.
My Approach towards the Proof:
Picking an edge $(X,Y)$ for maximal matching is wrong if there exist edges $(X,A)$ and $(Y, B)$. Clearly, picking $(X,Y)$ increases set of maximal matching by one, but we could have picked the other two edges and increase the cardinality by two.
In case there is only one edge $E$ having vertex $X$ as one of its end-point, we never come into the situation of missing out two edges when we include this one.
Best Answer
I think you have the correct idea. Maybe we can formalize it little more.
Let $G$ be a bipartite graph and $X$ and $Y$ are bipartitions of $V(G)$. Also let $x\in X$ be a vertex with degree $1$ and $e = xy$ be the edge incident to $x$. So, the edge that is connected to $x$ is connected to $y \in Y$. Then the claim should be there exists a maximal matching containing $e$ because we can find an example where there is a maximal matching does not contain $e$ with the same number of edges (but your claim is just to include $e$ to maximal matching safely so I don't think this is important here).
Now, let $M$ be the maximal matching with $|M|$ edges including $e$. Suppose for a contradiction that there exists a matching $M'$ with $|M'| \ge |M|+1$. Since $M$ is the maximal matching including $e$, $M'$ cannot include $e$. Therefore, only possibility for this matching is to include $ay$ as an edge where $a \in X$. Then we can remove $ay$ and add $e=xy$ to $M'$ without changing $|M'|$. But then our new matching is a matching that includes $e$ and with size $|M'| > |M|$, contradictory to the assumption that $M$ is the maximal matching including $e$.