[Math] Isomorphism of directed graphs

Question

Consider the following directed graphs:

One is obtained from the other by reversing the direction of all edges.

Are they isomorphic as directed graphs ?

On the one hand, I would answer: no because there is no pair of maps between the vertices and the edges respectively that preserves the adjacency relation.

On the other hand, if one forgets about the “labels'' on say, the edges, then the graphs are the same (just exchange the label 'e' with label 'f').

Answer 1

Your comment helps clarify the source of your confusion:

A graph morphism is a pair of maps between the respective set of vertices $p:V \to V$ and and between the respective set of edges $q:E \to E$. If I set $q(e) = f$, $q(f) = e$ and $q(l) = l$ then because of the adjacency relation, I have to set: $w = \text{initial vertex of } f = \text{initial vertex of } q(e) = p(\text{initial vertex of e})= p(v)$. So $p$ has to exchange $v$ and $w$. Now $q(l) = l$ so again by the adjacency relation one must have: $v = \text{initial of vertex of } l = \text{initial vertex of } q(l) = p(\text{initial vertex of } l) = p(v)$. Hence $p$ has to fix the vertex $v$. This is a contradiction.

You implicitly (and wrongly) assume that edge $f$ in the left graph is the same as edge $f$ in the right graph. The confusion is only due to name clash. Renaming should save you from this mistake.

If we named edges in the right graph $e'$ and $f'$, then instead of writing this statement (original):

$q(e) = f$, ... $w$ = initial vertex of $f$

we would have written this statement:

$q(e) = f'$, ... $w$ = initial vertex of $f$