# Do all equivalences in the 2-category of spans come from isomorphisms?

See my previous question What is the product in the 2-category of spans? for notation. In brief, $\mathcal S$ is a category with finite limits, $\operatorname{Span}(\mathcal S)$ is the 2-category whose 1-morphisms are diagrams in $\mathcal S$ of the form $\bullet \leftarrow\bullet \rightarrow \bullet$, and for want of a better name (is there one?) I call the functor $\{X\to Y\} \mapsto \{X = X \to Y\}$ "spanishization".

Suppose I have an equivalence in $\operatorname{Span}(\mathcal S)$ (a pair of 1-morphisms whose compositions are isomorphic to identities). Is it necessarily isomorphic to the spanishization of an isomorphism in $\mathcal S$?

• Overuse of "ish"! – Scott Morrison Jul 19 '10 at 21:32

Yes. Suppose we have spans $X_1 \stackrel{f_1}{\leftarrow} Y \stackrel{f_2}{\to} X_2$ and $X_2 \stackrel{g_2}{\leftarrow} Z \stackrel{g_1}{\to} X_1$ with an isomorphism $Y \times_{X_2} Z \stackrel{\omega}{\to} X_1$ such that $\omega = f_1 \circ \pi_1 = g_1 \circ \pi_2$, where $\pi_1$ and $\pi_2$ are the projections from $Y \times_{X_2} Z$ onto its factors. The morphism $f_1: Y \to X$ yields a 2-morphism from the $Y$ span to the spanishization of $f_2 \circ \pi_1 \circ \omega^{-1}: X_1 \to X_2$. The map $\pi_1 \circ \omega^{-1}: X_1 \to Y$ yields an inverse 2-morphism. Thus, the $Y$ span is isomorphic to the spanishization of $f_2 \circ \pi_1 \circ \omega^{-1}$.
The same argument using the other composite gives an isomorphism between the $Z$ span to the spanishization of $g_1 \circ p_1 \circ \beta^{-1}: X_2 \to X_1$, where $p_1: Z \times_{X_1} Y \to Z$ is the first projection, and $\beta: Z \times_{X_1} Y \to X_2$ is our isomorphism with the identity span. That the two morphisms we have constructed are inverse to each other in $\mathcal{S}$ follows from the fact that their spanishizations compose to something isomorphic to the identity span in both directions.