Biconnectivity of two subgraphs

Question

Let's say we have two graphs $G$ and $H$ with exactly two common vertices $u$ and $v$ and one common edge $uv$ and we know that graph $G \cup H$ is biconnected.

Is it true that $G$ and $H$ are biconnected too? If so, how can one show it?

Answer 1

This is true. To show that $G$ is biconnected take any three nodes $x,y,z$ in $G$. We want to show that there still exists a path from $x$ to $z$ in $G - y$.

Since $G\cup H$ is biconnected there exists a path $p$ from $x$ to $z$ in $G \cup H - y$. Now, we partition $p$ into three subpaths $p_1, p_2$ and $p_3$ with $p_1$ and $p_2$ connecting at some node $a$ and $p_2$ and $p_3$ at some node $b$ such that $a$ is the first node on $p$ which is in $H$ and $b$ the last such node. Clearly, we have $a,b \in \{u,v\}$.

If we have $a=b$, then $p_1, p_3$ is a path from $x$ to $z$ in $G-y$. Otherwise $p_1,\{u,v\},p_3$ is such a path. Thus, $x$ and $z$ are still connected.