If $G_1$ $\cong$ $G_2$ and $G_1$ has a cut edge $\Rightarrow$ $G_2$ has a cut edge For proving If $G$ $\cong$ $H$ and $G$ has a cut edge, e  $\Rightarrow$ $H$ has a cut edge, the way I am thinking about it is if I take 
$G = \langle V,E\rangle$ and let $G' =\langle V,E-e\rangle$.  Because $G$ $\cong$ $H$ $\exists f:E(G) \to E(H) $ that is bijective so for a function $g:E(G') \to E(H)-f(e) $ all of the isomorphic properties would still hold, which would imply that $f(e)$ would be a cut edge for $H$. 
I don't think this is a strong enough argument and it's not very well put together. How should I go about changing the logic to make it more precise and acceptable?
 A: It sounds like you have the right mental model. The property of "having a cut edge" is a "graph-theoretic property" in that it only depends on the "shape" of the graph, not on the particular identities of the vertices and edges. Removing $e$ from $G$ corresponds to removing $f(e)$ from $H$, and the resulting graphs $G'$ and $H'$ are still isomorphic to each other, so $H'$ is disconnected iff $G'$ is disconnected (because connectedness is a graph-theoretic property).
Here's one way of making the argument more formal. Suppose $f:V(G)\to V(H)$ is a graph isomorphism and $G$ has a cut edge $e=(v,w)$. We want to show that $f(e)$ (shorthand for $(f(v), f(w))$) is a cut edge of $H$. Suppose it's not. That is, $H$ remains connected after removing $f(e)$. This implies that there's a path $v_1, \dots, v_n$ in $H$ that connects $f(v)$ to $f(w)$ without using the edge $f(e)$. Since $f$ is a bijection, we have a corresponding sequence of vertices $f^{-1}(v_1),\dots,f^{-1}(v_n)$ in $G$, and since $f$ is a graph isomorphism, it preserves the existence (and non-existence) of edges between pairs of vertices, so this sequence is a path in $G$ that connects $v$ and $w$ without using $e$. This contradicts our assumption that $e$ is a cut edge.
