Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $\mathcal S$?

share|improve this question
3  
Overuse of "ish"! –  Scott Morrison Jul 19 '10 at 21:32

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer
    
In fact, more is true. By a similar argument, any morphism in Span(S) which has a right adjoint is (isomorphic to) the graph of a morphism in S. Since any equivalence can be improved to an adjoint equivalence, this implies that any equivalence in Span(S) is isomorphic to the graph of an isomorphism in S. (I call that functor "taking the graph of a morphism." "Spanishization" sounds to me like translating into Spanish. (-:) –  Mike Shulman Jul 20 '10 at 6:40
    
The Spanish pun goes back a few years golem.ph.utexas.edu/category/2007/07/…. –  David Corfield Jul 20 '10 at 9:00

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.