Show that any graph can be written as an edge disjoint union of its blocks Actually I am not very comfortable with using blocks, I understand the definition that it is a maximal $2$-connected graph, though.
My attempt Suppose not. Then there exists a maximal graph $G$ which cannot be written as an edge disjoint union of its blocks.
I cannot go further. I am not even sure why maximal exists, because the number of vertices of $G$ are not predetermined here. I am not even sure how can we write complete graph in such a fashion.
Please help
 A: I think we need maximality of the blocks rather than maximality of the graph here.
Assume there is a graph, which cannot be written as edge disjoint union of blocks, that is to say there are two distinct blocks $A,B$, which share a common edge $xy$. Without loss of generality(1) we can assume that $x$ has a neighbor $a$ in $A\setminus B$. As $y$ and $a$ lie in $A$ we can find an $ay$-path $p$ in $A$ avoiding $x$ and hence $xy$. But this means that $B \cup p \cup xa$ is a 2-connected subgraph strictly containing $B$, a contradiction.
(1) Take an edge $uv$ in the intersection and a vertex $w$ in $A\setminus B$. By connectivity we have a $vw$-path. Traveling along this path we find the first edge that leaves the intersection (it may go into $B\setminus A$ but that is fine: just swap the meaning of $A$ and $B$).
A: A block of a graph G is a maximal connected subgraph of G that has no articulation point(cut-point or cut-vertex). If G itself is connected and has no cut-vertex, then G is a block.Two or more blocks of a graph can meet at a single vertex only, which must necessarily be an articulation point of the graph.Hence, any graph can be written as an edge disjoint union of its blocks, because of their maximality.
We can prove this by contradiction.
Suppose we surmise a graph G which cannot be represented as an edge-disjoint union of blocks. For simplicity, let us consider that there are two different blocks X and Y which have a common edge uv. Without loss of generality, we are at liberty to assume that u has an adjacent vertex m in X. And v and m are vertices in X, therefore, there exists a path(P) between v and m not containing u. Evidently, P does not contain the edge uv. Also a path can be represented as a union of unique edges(k2 graphs, which are by definition blocks). Let S be the set of edges, in X-uv-um-P. And thus G can be represented by Y ∪ um ∪ P ∪ S which can be intuitively viewed as a union of different edge-disjoint blocks, which is a contradiction.
