Factor-critical Graph and Its Blocks

Question

I'm struggling to solve the statement: A Graph $G$ is factor-critical if and only if every block of $G$ is factor-critical.

For the if part, I tried to use the property of blocks that, two distinct block shares at most one vertex, thinking of deleting the common vertex and obtaining the perfect matching of each block without such vertex. But I stuck in that the union of such matching must not be a perfect matching of $G$.

For the converse, I thought about using the maximality of blocks, but idea does not follow easily.

It will be glad if someone give me a hind to solve this problem.

Answer 1

Here's something that looks valuable for both directions of the proof. An end-block of a graph is a block that contains only a single cut-vertex. If a graph has any cut-vertices at all, it contains at least two end-blocks. We can pull these off / add them on one by one. Here's the useful part (hover your mouse over to see hints):

If $G$ is a connected graph with an end-block $B$, and $v$ is the cut-vertex joining $B$ to the rest of $G$, then $G$ is factor-critical if and only if both of the induced subgraphs $G[B]$ and $G[G-(B-v)]$ are factor critical. In other words, we can pull an end-block off a factor critical graph and obtain a factor critical graph with fewer blocks.

You still need to prove the hint, but it should make both directions easier.

Another hint:

Say we have two connected graphs $H$ and $K$ with a common vertex $v$ such that $H\cap K = \{v\}$. Let $M_H$ be a perfect matching of $H$, and $M_K$ a perfect matching of $K-v$, then $M_H\cup M_K$ is a perfect matching of $H\cup K$.

And a fact to keep in mind for proving the first hint:

Any graph with a perfect matching has even order. Any factor critical graph has odd order.

This is all still far from a proof, and the 'main idea' behind this result comes up in proving the first hint, so let me know if you'd like something more concrete.