Reading through the Category Theory Wikibook, I came across the following exercise:
(Harder.) If we add another morphism to the above example, it fails to be a category. Why? Hint: think about associativity of the composition operation.
I haven't been able to figure out why the diagram given is not a category. The circles are the objects, the arrows are the morphisms, and there is a clear notion of composition.
Going through the category laws (in reverse order):
1. There is an identity element for every object ($id_A$ and $id_B$).
2. The set of morphisms are closed under composition: $f \circ g > = id_A$, $h \circ g = id_A$, $g \circ f = id_B$, $g \circ h = id_B$, and any composition involving an identity morphism is clearly (I think) closed. Compositions like $f \circ h$ are meaningless.
3. Coming to the hint in the prompt, and trying to prove associativity, $h \circ (g \circ f) = g \circ id_B = g$, whereas $(h > \circ g) \circ f = id_B \circ f = f$, so $h \circ (g \circ f) \neq (h > \circ g) \circ f$.
This is where I get a little confused, though, since $f$ and $h$ are in some ways "the same", since A and B are isomorphic (right?). I haven't figured out exactly why I'm confused about this, but it seems like this exercise means that there can only ever be a single morphism between two objects in a category, which is clearly nonsense. What am I missing in all this?