Block containment and monotonicity under subgraphs

Let $C$ be a 2-connected subgraph of $L$. The finite, nonempty family of 2-connected subgraphs of $L$ containing $C$ has a member $M$ maximal under subgraph containment: otherwise a strictly increasing chain could be extended indefinitely within that finite family. Any 2-connected subgraph properly containing $M$ would also contain $C$, so $M$ is maximal among all 2-connected subgraphs of $L$.

Every edge $e=uv$ of $L$ lies in a nontrivial block. If $e$ is a bridge, its endpoint subgraph is such a block. Otherwise $u$ and $v$ remain in the same component of $L-e$: if they were in different components, adding $e$ back would merge exactly their two components and deletion of $e$ would increase the component count. A path from $u$ to $v$ in $L-e$, together with $e$, forms a simple cycle. This cycle is 2-connected, so step 1.1 puts it, and hence $e$, in a maximal 2-connected subgraph of $L$.

If $J$ is edgeless, then $b(J)=1\le b(L)$: the right side is $1$ when $L$ is edgeless and at least $2$ otherwise, by step 2.1. Suppose instead that $J$ has an edge, and choose a nontrivial block $B$ of order $b(J)$. If $B$ is 2-connected, its image under $f$ is a 2-connected subgraph of $L$: the isomorphism preserves paths and vertex deletion, so connectivity and connectivity after each vertex deletion are preserved. By step 1.1 its image lies in a nontrivial block of $L$. If $B$ is a bridge-edge subgraph of $J$, its image edge lies in a nontrivial block of $L$ by step 2.1, whether or not that image edge is a bridge in $L$. In either case injectivity gives $\mathrm{\mid }V(B)\mathrm{\mid }\le b(L)$, proving $b(J)\le b(L)$.

If a 2-connected graph occurs as a subgraph of $L$, step 1.1 contains it in a maximal 2-connected subgraph of $L$. Its order is therefore at most $b(L)$, proving the final assertion.