# Are adjoints closed under pushouts?

The category $$PrL$$ of locally presentable categories has all colimits. In particular, if $$A_1 \leftarrow A_0 \rightarrow A_2$$ is a diagram of presentable categories, with left adjoint functors between them, then the pushout $$A_1 \coprod_{A_0} A_2$$ in $$PrL$$ exists and the pushout functors $$A_1 \rightarrow A_1 \coprod_{A_0} A_2, A_2 \rightarrow A_1 \coprod_{A_0} A_2$$ are morphisms in $$PrL$$ and hence are left adjoints.

I am wondering whether this generalizes to categories that are not locally presentable. Concretely, if $$A_1 \leftarrow A_0 \rightarrow A_2$$ is a diagram of categories (maybe not presentable) with left adjoint functors between them, we can form the pushout $$A_1 \coprod_{A_0} A_2$$ in $$Cat$$.

Question: Are the pushout functors $$A_1 \rightarrow A_1 \coprod_{A_0} A_2$$ and $$A_2 \rightarrow A_1 \coprod_{A_0} A_2$$ also left adjoints?

If the answer is no in general, are there additional conditions on the diagram $$A_1 \leftarrow A_0 \rightarrow A_2$$ that imply that the answer is yes?

This question Pushouts in the category of adjunctions is relevant but I think does not answer my question since I am not interested in pushouts in the category of adjunctions $$Cat^{Adj}$$, which I think does not have all pushouts by the answers there.

Remark: I am thinking about this question for $$\infty$$-categories and $$Cat$$ means that $$\infty$$-category of $$\infty$$-categories. It might be useful to consider $$Cat$$ as an $$(\infty, 2)$$-category but I am not sure.

• The answer is presumably “no”. Even if it were “yes”, this would not be a generalization of the fact in $Pr^L$ because the forgetful functor $Pr^L \to Cat$ does not preserve pushouts. Commented May 28 at 14:53
• To precise Tim's Comment : pushout in the category $Pr^L$ are computed as pullback in Pr^R, which themselve are computed as pullback in Set. The "generaization" of this fact about $Pr^L$ would be a question about pullback of right adjoint functors instead. Commented May 28 at 16:56
• It is much harder than I thought it would be to come up with a counterexample! Commented May 29 at 15:00

No. Here is a counterexample. Let $$n$$ denote the linear order with $$n$$ elements. The pushout of the left adjoints $$1 \leftarrow 2 \to 3$$ is the walking retract $$R$$ where the second map is the inclusion of the endpoints. The functor $$3\to R$$ is not a left adjoint.

• Note that $1,2,3$ are in $PrL$. $R$ is not (it doesn’t have binary products, although it does have a terminal object and equalizers I believe). Commented Jun 4 at 19:44
• Interesting! I wonder what happens if we assume that the pushout is in PrL; then do the morphisms have adjoints? Commented Jun 6 at 10:50