Let $G$ be a graph and let $v$ be a vertex of $G$. Let $\alpha'(G)$ denote the size of maximaum matchings in $G$.
My question is the following:
Question 1. What is the maximum value of $\alpha'(G) - \alpha'(G-v)$, where $G-v$ is the graph obtained from $G$ by removing vertex $v$?
I think $\alpha'(G)-\alpha'(G-v)\le 1$ is correct. Because a vertex can be saturated by at most one edge in a (maximum) matching.
Here is an example I am considering. If we remove vertex $v$, the matching number decreases by 1. If we remove vertex $u$, the matching number remains unchanged.
What happens if we extend the situation of removing one vertex to removing a vertex set?
Question 2. What is the maximum value of $\alpha'(G) - \alpha'(G-S)$, where $G-S$ is the graph obtained from $G$ by removing vertex set $S$ with $|S|=k$?
Will we obtain an expression in terms of $k$? How about the simple case where $k=2$?
I think I have got some ideas of your question. Let's consider the general situation of the question which is, the maximum value of α'(G)-α'(G-S) where S is a set of vertex and |S|=k.
Firstly, as you had said before, since a vertex can only be saturated by one edge, it is obvious that α'(G) - α'(G-S)$\leq$k
Whether α'(G) - α'(G-S) can be tightly equals to k is a more intriguing question.
If the graph G is a acyclic graph, I can prove that the limit of k is tight.
definition: if a vertex have only one edge connected to other vertexes, we call it a leaf
prove: We assume that all leaves aren't matched, choose one leaf randomly and let's call it u and the vertex it connected to is v.
if v is not matched to any other vertex, than let u match with v, we can obtain another mathcing without disturbing the the previous matching.
if v is matched with another vertex, let's say, w. Than we can cancel the matching between w and v and let u to be connected with v which won't influence the result. So there must be a maximum matching where u is matched with v.
Q.E.D.
prove: According to lemma 1, there must be at least one leaf that is in the matching. Let's call it u and the vertex it connected to is v. If we delete v, it is obvious that u is useless.(since it isn't connected to any vertex at all!)
If α'(G) = α'(G-v), than it means we can have a maximum matching method whose size is as big as the maximum matching after we deleted two adjacent vertexes.
This as quiet a contradiction because if we let the two vertexes matched, α'(G) will definitely raise.
So we proved that α'(G) $\neq$ α'(G-v), since we have α'(G) - α'(G-S)$\leq$1, we can get α'(G) - α'(G-S) = 1
Q.E.D.
We find that the graph after we deleted v is still a acyclic graph, which means we can do the former operation recursively with the help of Lemma 1 and Lemma 2 until α'(G-S) reach zero.
So we have solved the situation when G is a acyclic graph. But when there are circles in G, I'm not sure whether there is a strict condition for α'(G) - α'(G-S) = k or not, but I'm sure not all of them can.
Just consider a graph that contains only three vertexes, let's call them u, v and w.
there are three edges between u and v, u and w, v and w respectively, than it can be easy proved that you can't make α'(G) decrease just by deleting one vertex (the other two can still make a match).