Max Flow Min Cut - Prove that $e$ crosses some minimal cut

Question

I already asked about the opposite direction but I'm really confused about it, so I'd like to get some help please:

Let's assume we have a flow network $G$ and some edge $e$. Now, Let's assume that $G'$ is the same graph, only without the edge $e$. Also, let's assume that $|f_G|=|f_{G'}|+ c(e)$. Prove that $e$ crosses some minimal-cut.

So basically I thought about looking at some minimal cut $(S,T)$ of $G'$. Then, we look at at some minimal cut $(\hat{S} ,\hat{T})$ of $G$.

By assumption, $$C(\hat{S} ,\hat{T}) = |f_G| = |f_{G'}| + c(e) = C(S,T) + c(e)$$

Therefore, $(S,T)$ is the minimal cut for $G$ where $e$ crosses it.

I'm not 100% sure about this.

Could you please help me understand better this?

Thanks!

Answer 1

As we have the min-cut $C(S',T')$ in $G'$. Denote the cut edges by $E'$ which has weight $|f_{G'}|$. If we adding back $e$ to $G'$, then $E'$ cannot be a edge cut anymore, since otherwise $|f_G|\leq|f_{G'}|$, here we actually assume that $c(e)>0$. But $E'+e$ must be a edge cut in $G$, and it has weight $|f_{G'}|+c(e)=|f_{G}|$, which means that it is a min-cut of $G$.