The proof you've provided establishes that if $G$ is a connected graph and $g$ is an isomorphism, then $G'$ (the graph obtained by applying $g$ to $G$) is also connected. The next part of the proof asks to prove the converse: if $G'$ is connected, then $G$ is connected.

Here's how you can extend the proof:

Suppose $G'$ is connected.
Let $x, y \in V$ (the vertex set of $G$). Since $g$ is an isomorphism, there exist vertices $u, v \in V'$ (the vertex set of $G'$) such that $g^{-1}(u) = x$ and $g^{-1}(v) = y$.

Since $G'$ is connected, there exists a walk in $G'$ from $u$ to $v$:
$u = u_0, u_1, \dots, u_n = v,$ with edges $e_i = \{ u_i, u_{i+1} \} \in E'$ for $i = 0, 1, \dots, n-1$.

Because $g$ is an isomorphism, its inverse $g^{-1}$ maps vertices and edges of $G'$ to vertices and edges of $G$:

• $g^{-1}(u_i) \in V$,
• $g^{-1}(e_i) = \{ g^{-1}(u_i), g^{-1}(u_{i+1}) \} \in E$.

Therefore, there is a walk in $G$ from $g^{-1}(u) = x$ to $g^{-1}(v) = y$:
$x = g^{-1}(u_0), g^{-1}(e_0), g^{-1}(u_1), \dots, g^{-1}(e_{n-1}), g^{-1}(u_n) = y.$

This proves that $G$ is connected.

Conclusion: Since we have shown both directions (if $G$ is connected, $G'$ is connected, and if $G'$ is connected, $G$ is connected), we conclude that $G$ and $G'$ are connected if and only if $g$ is an isomorphism.

Let me know if you would like further clarification! Here are some related questions you might explore:

1. What properties of isomorphisms ensure that connectivity is preserved?
2. Can an isomorphism exist if one graph is connected and the other is not?
3. How does this proof generalize to directed graphs?
4. What additional properties, besides connectivity, are preserved under isomorphism?
5. How would you prove connectivity for infinite graphs under isomorphism?

Tip: When dealing with graph theory proofs, always verify that each step logically connects the properties of the graph and the definition of the operation (like isomorphism).