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:
- $f: A \longrightarrow B$ 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)
- 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 $f\circ I_A \cong f$.