If $u$ and $w$ belongs to the same connected components, does there exist any $u-w$ path containing $v$?

Question

If $u$ and $w$ belongs to the same connected components, does there exist any $u-w$ path containing $v$?

Answer 1

If $v$ is a cut vertex of $G$, then $G-v$ is disconnected and has at least two components , $G_1$ and $G_2$. Take $u \in G_1$ and $w \in G_2$.

You ask, “what happens if $u$ and $w$ lie in the same connected component of $G-v$?”

If $u$ and $w$ were in the same component of $G-v$, let's say $G_1$, there would be a path in $G_1$ connecting $u$ to $w$. This path would not contain $v$, because $G_1$ is a component of $G-v$. It might happen that some paths in $G$ connecting $u$ to $w$ contain $v$, but not all of them.

But notice that $u$ and $w$ aren't arbitrary vertices—far from it. We only need to show that such vertices exist somewhere in $G$. So we carefully require that $u$ and $w$ come from different components of $G-v$. Now we know there is a path connecting $u$ to $w$ in $G$ (because $G$ is connected), but there is no path connecting $u$ to $w$ in $G-v$ (because $G-v$ is not connected, and $u$ and $w$ are in different components), so every path from $u$ to $w$ in $G$ must not be a path in $G-v$. That is, every path from $u$ to $w$ in $G$ must contain $v$.

Answer 2

Not in $G_1 \cup G_2$, since $v$ is contained neither in $G_1$ nor in $G_2$. Indeed, $G_1 \cup G_2 = G \setminus \{v\}$. However, there exists a path in $G$ because $G$ is connected.