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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.

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    $\begingroup$ 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. $\endgroup$
    – Tim Campion
    Commented May 28 at 14:53
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    $\begingroup$ 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. $\endgroup$
    – Simon Henry
    Commented May 28 at 16:56
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    $\begingroup$ It is much harder than I thought it would be to come up with a counterexample! $\endgroup$
    – Chris Schommer-Pries
    Commented May 29 at 15:00

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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.

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    $\begingroup$ 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). $\endgroup$
    – Tim Campion
    Commented Jun 4 at 19:44
  • $\begingroup$ Interesting! I wonder what happens if we assume that the pushout is in PrL; then do the morphisms have adjoints? $\endgroup$
    – user39598
    Commented Jun 6 at 10:50

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