Span Categories

Question: It seems to me, the bicategory of spans, with canonical choice of units, has strict units, so it is not a general bicategory. Am I right?

Below is the definition:

Definition: Consider a small category $C$ which has pullbacks, a bicategory of spans $D$ consists of:

1. $Obj(D)=Obj(C)$
2. $f: A \longrightarrow B$ is a pair of morphisms from an arbitrary object $X$, $f=(f_A:X \longrightarrow A, f_B:X \longrightarrow B)$ such that the composition of two morphisms is the pullback of them. (We fix compositions of each pair of morphism to avoid weak equality)
3. 2-cell $\alpha: f \longrightarrow g$ is $\alpha: X \longrightarrow Y$ such that $f=(f_A:X \longrightarrow A, f_B:X \longrightarrow B)$ and $g=(g_A:X \longrightarrow A, g_B:X \longrightarrow B)$.

Identity is a pair of identity morphisms of the original category, so $I_A=(id_A, id_A)$. Therefore, $\lambda: f \circ I_A \longrightarrow f$. The canonical pullback of $f \circ I_A$ is $(X, id_X, f_A)$ so : $$f\circ I_A =(id_A \circ f_A, f_B \circ id_X)$$ But $id_A \circ f_A = f_A$ and it is equality because of the composition defintion in the main category $C$. So $f\circ I_A =f$ not $f\circ I_A \cong f$.

• $u=v$ implies $u \cong v$.... – Excluded and Offended Aug 13 '18 at 12:01
• @MobiusKnot: Yes, that's right, it's a stirct identity. – Berci Aug 13 '18 at 20:50

It's certainly possible to choose the pullback $B\times_A A$ to be equal to $B$, but that's unlikely to be implied by any reasonable choice of all pullbacks: for instance, this certainly doesn't hold for usual constructions of pullbacks in the category of sets. So I would say the bicategory of spans doesn't naturally have strict units; you can force it to have strict units, but then it's also equivalent to a fully strict 2-category, so that's not exactly surprising.