Cospans for finite sets problem

Let $\mathcal{C}$ be the category formed by three objects $A, B, C$ and the following diagram: $A \to C \leftarrow B$ (let $\alpha$ be the map from $A$ to $C$, and $\beta$ the map from $B$ to $C$).

It is then possible to form the category $\mathcal{C}'=Cospan(\mathcal{C})$. From what I understood, $\mathcal{C}'$ has an invertible morphism between $A$ and $B$.

My question is the following: if $A, B, C$ are sets, how can there be an isomorphism between $A$ and $B$, as $\alpha(A)$ might not be equal to $\beta(B)$ ? Obviously there is something wrong in my reasoning so any help is appreciated.

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You haven't explained what $\mathcal{C}'$ is at all. Under the most obvious interpretation I can think of, $A$ and $B$ are not literally objects of $\mathcal{C}'$, and even if we relax the interpretation a little, they are not isomorphic. –  Zhen Lin Jul 20 '12 at 17:12
@Zhen: The cospan category $\mathcal{C}'$ is defined e.g. in this paper. –  joriki Jul 20 '12 at 17:20
@joriki: That's the bicategory of cospans, which doesn't make sense here since $\mathcal{C}$ doesn't have pushouts. (Moreover, a bicategory is not a category.) –  Zhen Lin Jul 20 '12 at 17:28
@Zhen: I'm quite a bit out of my depth here, so I may very well be wrong, but I was under the impression that the cospan category is a double category, not a bicategory, and that a double category can be viewed as a plain category by considering either the horizontal morphisms or the vertical morphisms. Also, if the objects are sets, then we can take the pushout to be the disjoint union, so it does seem to make sense to consider the existence of pushouts implied. –  joriki Jul 20 '12 at 17:38
Composition of cospans is given by pushouts, which is not strictly associative in most categories. But as it turns out, $\mathcal{C}$ has strictly associative pushouts. (What it lacks is pullbacks, my mistake!) So in this case the cospan bicategory is an honest category, extremely unusually. Nonetheless, $A$ and $B$ are not isomorphic in this category. –  Zhen Lin Jul 20 '12 at 17:55