Help with proof that the union of two undirected cycle graphs is a cycle graph (with two edge deletions) I am seeking advice on how to prove something. Apologies if my terminology is incorrect: I am not a mathematician.
Let $G_1$ and $G_2$ be undirected cycle graphs with edges $E_{G1}$ and $E_{G2}$ respectively. 
Let $e_{G1} = (a,b)$ and $e_{G2} = (x,y)$ be arbitrary edges from $E_{G1}$ and $E_{G2}$ respectively. 
Remove these edges from their respective graphs.
For every edge in $G_1$, substitute $a$ for $x$ and $b$ for $y$.
Let $G_3 = G_1 \cup G_2$.
$G_3$ is a undirected cycle graph.
I would like to prove this, but I have literally no idea how to do this. Can anyone offer any advice?
Thanks
 A: This is the type of result that's much easier to intuit than it is to rigorously prove. For a rigorous proof, it's probably best to enumerate the vertices and edges of each graph:
Let $G_1$ and $G_2$ be cycles with:
$$V(G_1)=\{u_1, u_2, \dots, u_{m}\}$$ 
$$E(G_1) = \{u_1 u_2, u_2 u_3, \dots, u_{m-1} u_{m}, u_{m} u_1\}$$
$$V(G_2)=\{v_1, v_2, \dots, v_{n}\}$$ 
$$E(G_2) = \{v_1 v_2, v_2 v_3, \dots,v_{n-1} v_{n},v_{n} v_1\}$$
You may assume that the edge to be deleted from $G_1$ is $u_m u_1$, and the edge to be deleted from $G_2$ is $v_1 v_n$ because of symmetry (otherwise you could relabel the vertices to make it so). Then the graph $G$ obtained by deleting $u_m u_1$ from $G_1$ and $v_n v_1$ from $G_2$ and identifying vertex $u_1$ with vertex $v_n$ (calling the new vertex $u_1$) and vertex $u_m$ with vertex $v_1$ (calling the new vertex $u_m$) is the graph you are looking for. Now we have:
$$V(G) = \{u_1, u_2, \dots, u_m, v_2, v_3, \dots, v_{n-1}\}$$
$$E(G) = \{u_1 u_2, u_2 u_3, \dots, u_{m-1} u_m, u_m v_2, v_2 v_3, v_3 v_4, \dots, v_{n-2} v_{n-1}, v_{n-1} u_1\}$$ 
But this is exactly a cycle on $m+n-2$ vertices.
