See my previous question [What is the product in the 2-category of spans?](https://mathoverflow.net/questions/32526/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".



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> 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 $\mathcal S$?
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Do all equivalences in the 2-category of spans come from isomorphisms?
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.


