# Max-flow of undirected graph

I know that a max flow problem for undirected graphs has a standard reduction to directed graphs in order to apply Ford–Fulkerson algorithm and getting a feasible solution.

Here's my doubt. An undirected edge $$(u,v)$$ with capacity $$1$$, will correspond to two directed edges $$(u,v), (v,u)$$ both with capacity $$1$$. However, the solution cannot use both edges since I would lose the correspondence with the original problem.

What am I missing?

If a flow somehow uses both $$(u, v)$$ and $$(v, u)$$, it is equivalent to another flow that only uses one of them. So if $$f_{uv} \ge f_{vu} > 0$$, treat it as the equivalent flow $$\hat{f}$$ where $$\hat{f}_{a} = f_a$$ for $$a\notin\lbrace (u, v), (v, u) \rbrace$$, $$\hat{f}_{vu} = 0$$ and $$\hat{f}_{uv} = f_{uv} - f_{vu}$$.
I'm tempted to say that this will not happen -- that a maximal flow will always use at most one of $$(u,v)$$ and $$(v,u)$$ -- but in some cases that might depend on the efficiency (or dopiness) of the algorithm being used to compute it.