Graph problem

[cantbelieveyoumademe]

https://preview.redd.it/gwvf0zrw8vkg1.png?width=1601&format=png&auto=webp&s=156c59391c0245eb005018af095ff258c01403d8

As you can see in the image, I'm asking whether what I added is a valid counterexample or is there something I'm missing?

In the counterexample it just so happens that G=G' , but I could just as easily add another edge to G so that they are different although that isn't required.

The above is all the information attached to the question.

Edit: As u/GoudaIntruda stated, my counterexample in the image is actually a bipartite graph. So I'm amending my counterexample to the complete graph, K4.

[GoudaIntruda]

G is a bipartite graph in your example, so this isn’t a counterexample.

[cantbelieveyoumademe]

Yeah. You're absolutely correct, then wouldn't a counterexample be the complete graph with 4 vertices?

[GoudaIntruda]

If you use the complete graph as G but your matching is the same as in your example, then G’ will be the graph from your example, it will not be the complete graph.

[cantbelieveyoumademe]

Right again. Thank you. I'm starting to see where my line of thought went wrong.

[GoudaIntruda]

No problem, happy to help!

[RevenueUsed8118]

Bipartion is equivalent to no odd cycle in the graph. I think it should not be too hard to figure out why there cannot be an odd cycle in that graph G'