Let $G=(V,E)$ be a simple, undirected graph. A *matching* is a set $M\subseteq E$ consisting of pairwise disjoint edges. We say $M$ is *maximal* if it is maximal amongst all matchings in $G$ with respect to $\subseteq$.


Given $n\in\mathbb{N}$ is there a graph $G$ such that $\chi(G)\geq n$ and every two maximal matchings have non-empty intersection?



Graphs in which all maximal matchings intersect
Affirmative: take $G$ to be the disjoint union of $K\_n$ and $K\_2$, then the $K\_n$ forces $\chi(G)\geq n$, while any maximal matching contains the $K\_2$ edge.


OK, that was cheating. I suppose you want a connected example.
I don't know the answer there, but can give a negative answer if you replace the condition $\chi(G)\geq n$ with $|G|\geq n$ and minimum degree of a vertex is at least $|G|/2$. For in this case it is known that $G$ has at least $\lfloor 5n/224\rfloor$ edge-disjoint Hamiltonian cycles (Nash-Williams). So if $n$ is even and at least $90$, then $G$ will have at least $2$ edge-disjoint Hamiltonian cycles. If you take every other edge in an even-length Hamiltonian cycle you will get a maximal matching. Since you have at least $2$ edge-disjoint Hamiltonian cycles, you end up with two edge-disjoint maximal matchings.


**Edit.** I guess you only need one even-length Hamiltonian cycle to get $2$ edge-disjoint maximal matchings: for one take every other edge in the Hamiltonian cycle and for the other take edges of the cycle not used in the first matching.


