Let $\mathcal A$ be the category whose objects are all categories, and whose morphisms from a category $\mathcal C$ to a category $\mathcal D$ are all adjunctions $F \dashv G$ with $F : \mathcal C \to \mathcal D$ (and $G : \mathcal D \to \mathcal C$). Does every span in $\mathcal A$ have a pushout, and, if yes, what does such a pushout look like?

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    $\begingroup$ Since this is obviously a 2-categorical situation, you should specify the higher structure to get a relevant answer. I think perhaps the setting of the double category of profunctors could be useful here. $\endgroup$ – Adam Gal Jan 24 '14 at 6:49
  • $\begingroup$ So I have to specify what morphisms between adjunctions should be. I think that the canonical answer would be that a morphisms from $F \dashv G$ to $F' \dashv G'$ is a pair of natural transformations $\varphi : F \to F'$ and $\gamma : G \to G'$ such that $\varepsilon' \circ (\varphi \bullet \gamma) = \varepsilon : F \bullet G \to \mathrm{Id}$ and $(\gamma \bullet \varphi) \circ \eta = \eta' : \mathrm{Id} \to G' \bullet F'$. However I do not know how pushouts in 2-categories are defined. Any hints? $\endgroup$ – Wolfgang Jeltsch Feb 17 '14 at 16:42

$ \newcommand{\Hom}{\mathrm{Hom}} \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}$ In general, this is false. Let $G_1$ and $G_2$ be two groups and consider the two free/forgetful adjunctions between finite sets and finite sets with an action of $G_i$. The pushout should equal finite sets with an action of the free product $G_1 * G_2$, but it cannot because the free functor takes you to infinite sets. If $H$ is any finite quotient of $G_1 * G_2$, then $H$-sets forms a commutative diagram in the category of adjunctions. If there were a universal adjunction then the unit of the universal adjoint functor applied to a free $G_1$-set would produce a $G_1$-set that is at least as large as the free object on one generator in any of these categories, which is a free $H$-set, and can grow arbitrarily large, so this is impossible.

However, this is correct as long as all categories involved have filtered colimits and the right adjoints commute with them. Daniel Schappi points out a simple argument for this in the comments, but my argument below could serve as an explicit construction.

I am not sure what happens if other colimits are missing or if the colimits exist but the functors do not commute with them.

My construction:

Tt might be helpful to do what I did when thinking of it and view the span in the category of adjunctions as consisting of the free/forgetful adjunctions $\mathbf{Set} \to \mathbf{Ring}$ and $\mathbf{Set} \to \mathbf{Group}$. Then the final category we are constructing is just sets with unrelated group and ring structures, and the main difficulty is categorically defining the free adjoint to the forgetful functor from ring-groups to rings or groups.

Let $F_1: \mathcal C \to \mathcal D_1$ and $G_1: \mathcal D_1 \to \mathcal C$ be one adjunction, and $F_2: \mathcal C \to \mathcal D_2$ and $G_2: \mathcal D_2 \to \mathcal C$ be the other.

Let $\mathcal B$ be the category of pairs of objects $X \in D_1, Y \in D_2$ with an isomorphism between $G_1 X$ and $G_2 Y$. There are natural projection functors $P_1: \mathcal B \to \mathcal D_1$ and $P_2: \mathcal B \to \mathcal D_2$. We will construct adjoints $Q_1: \mathcal D_1 \to \mathcal B$ and $Q_2 : \mathcal D_2 \to \mathcal B$.

We need an adjunction of the form $\Hom_{\mathcal B} (Q_1 X_0, (X,Y) ) = \Hom_{\mathcal D_1} (X_0,X)$. We have a natural map:

$\Hom_{\mathcal D_1}(X_0,X) \to \Hom_{\mathcal C} ( G_1X_0,G_1X) = \Hom_{\mathcal C}(G_1X_0,G_2 Y) \to \Hom_{\mathcal D_2} (F_2G_1 X_0, F_2G_2 Y) \to \Hom_{\mathcal D_2} (F_2 G_1 X_0, Y)$

Let $Y_1 = F_2G_1 X_0$. Then by the same logic we have a natural map $\Hom_{\mathcal D_1}(X_0,X) \to \Hom_{\mathcal D_1} ( F_1 G_2 Y_1, X)$. But we have something slightly more refined. $F_1 G_2 Y_1 = F_ 1 G_2 F_2 G_1 X_0$. Every element of $\Hom_{\mathcal D_1}(X_0,X)$ gives a commutative diagram:

$$ \begin{array}{c} F_1G_1X_0 & \ra{} & X_0 \\ \da{} & & \da{} \\ F_1G_2F_2 G_1 X_0 & \ra{} & X \end{array} $$

Call $X_2$ the limit of the top-left three-quarters of this diagram. Every map $X_0 \to X$ gives a natural map $X_2 \to X$. Then let $Y_3$ be the limit of the corresponding diagram starting at $Y_1$. Every map $X_0 \to X$ gives a corresponding map $Y_3 \to Y$.

Define $Y_4$, $X_5$, etc., similarly. Let $X_{\infty} = \lim_{n\to\infty} X_n$ and let $Y_\infty = \lim_{n\to\infty} Y_n$. Then $G_1 X_\infty = G_2 Y_\infty$, because they are both equal to the limit of the sequence $G_1 X_0 \to G_2 Y_1 \to G_1 X_2 \to G_2 Y_3 \to \dots$, because right adjoints preserve limits and the limit of a subsequence is the limit of the sequence. So $(X_{\infty}, Y_{\infty})$ is an element of $\mathcal B$, and we have a natural map $\Hom _{\mathcal D_1}(X_0,X)= \Hom_ {\mathcal B} ( ( X_\infty, Y_\infty), (X,Y))$. To see this is a bijection note that an element on the right side is just a sequence of maps $X_n \to X$, and $Y_m \to Y$. We need to show that all these maps are determined by the map $X_0 \to X$. They satisfy the consistency condition that the families of map $G_1 X_n \to G_1 X$, $G_2 Y_2 \to G_2 Y$ agree.

So for instance we have the unit map $G_1 X_0 \to G_2 F_2 G_1 X_0 = G_2 Y_1$. The map $Y_1 \to Y$ is the unique map such that the functor $F_2$ applied with it, composed with this unit, is the map $G_1 X_0 \to G_1 X=G_2 Y$. In other words it is exactly the map $F_2 G_1 X_0 \to Y$ one obtains by the adjunction. Iterating the process, we see that the maps from $X_2$, $Y_3$, etc. are also so determined.

So define $Q_1 X_1 = (X_{\infty}, Y_{\infty})$. Define $Q_2$ similarly. This completes our span to a commutative square. Is it universal? It clearly is one way. If $\mathcal B'$ is any category with functors $P_1': \mathcal B' \to \mathcal D_1$ and $P_2': \mathcal B' \to \mathcal D_2$, we obtain a unique map $\mathcal B' \to \mathcal B$ by definition. So we only need to construct an adjoint of that map. We do this as follows. Given $X,Y$, and an isomorphism $G_1X = G_2Y$, we form in $\mathcal B'$ the span

$$ \begin{array}{c} Q_1' F_1G_1 X & \ra{} & Q_1' X \\ \da{} & \\ Q2' Y & \end{array} $$

with the bottom arrow coming from the identification $Q_1'F_1G_1 X = G_2' F_2 G_1 X = Q_2' F_2 G_2 Y$.

If the pushout of this span in $\mathcal B'$ exists, we can use it to construct an adjoint functor. A $\Hom$ from this to an element $Z \in B'$ consists of a $\Hom(Q_1' X, Z)$ plus a $\Hom(Q_2'Y, Z)$ that agree on $\Hom(Q_1'F_1 G_1 X, Z)$. By adjunction, this is the same as a $\Hom(X, P_1'Z)$ plus a $\Hom(Y, P_1'Z)$ that agree on $\Hom(G_1X, G_1 P_1'Z)$. This is exactly a an element of $\Hom_B((X,Y), (P_1'Z,P_2'Z))$.

  • $\begingroup$ Small typo: "Then by the same logic we have a natural map $Hom_{\mathcal D_1}(X_0,X) \to Hom_{\mathcal D_1} ( F_2 G_1 Y_1, X)$" - $F_1 G_2 Y_1$ instead of $F_2 G_1 Y_1$. $\endgroup$ – Oskar Jan 25 '14 at 10:15
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    $\begingroup$ I don't think this works quite as intended. First of all, you often talk about limits when you really mean colimits (both in the definition of $X_i$ and $X_{\infty}$). But right adjoints need not preserve colimits (not even filtered colimits in general), so your argument that $G_1 X_{\infty}=G_2 Y_{\infty}$ does not work. $\endgroup$ – Daniel Schäppi Jan 25 '14 at 17:35
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    $\begingroup$ On the other hand, if you consider the 2-category of say lfp categories and adjunctions whose right adjoint does preserve filtered colimits, then you do get the desired pushouts. This 2-category is equivalent to finitely cocomplete categories with functors preserving finite colimits, and this is 2-monadic over the category of small categories. Since the 2-monad in question has rank, this 2-category has all bicolimits. I think it would be interesting to see if there are counterexamples if the categories involved do not have (many) colimits. $\endgroup$ – Daniel Schäppi Jan 25 '14 at 17:41
  • $\begingroup$ Good point. What are good examples of categories without filtered colimits? I don't know any way to find them other than enumerating all the categories I know and checking if they have filtered colimits, which seems tiresome. $\endgroup$ – Will Sawin Jan 25 '14 at 20:04

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