# Perfect matching in line graph

I am given a graph $T$ with an odd number greater than or equal to 3 of vertices. Its line graph $L(T)$ has exactly one perfect matching.

I need to prove that if we remove any vertex from $T$, the number of even connected components in $T$ will be even.

I need an idea on how to start on this. How exactly does existence of the perfect matching in $L(T)$ influence the original graph?

This is not true. Consider $$T=3P_3$$. It has an odd number (9) of vertices.
The line graph $$L(T)=3P_2$$, which has exactly one perfect matching.
Nevertheless, if you remove an endpoint $$v$$ from one of the paths in $$T$$ you are left with $$T-v$$ which has an odd number (1) of even components.
It is also not true if $$T$$ is connected. Consider the claw $$K_{1,3}$$ and subdivide each edge. Verify that it has an odd number of vertices and that its line graph has a unique perfect matching. Now if you remove the central vertex you are left with three components of size $$2$$.