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I know that for a given category $\mathcal C$, the diagonal functor $\Delta : \mathcal{C \to C \times C}$ has the product and coproduct functors on $\mathcal C$ (if they exist) as its right and left adjoints respectively. But what about the codiagonal functor $\nabla : \mathcal{C + C \to C}$?

My thought is that $\nabla$ can't have adjoints because any functor $\mathcal{F : C \to C + C}$ has to send each object $X \in obj(\mathcal C)$ to one or the other 'side' of $\mathcal{C + C}$ (call them the 'left' and 'right' sides). Say $X$ is sent to the left side: then taking the right-hand copy of $X$ from $\mathcal{C + C}$ and applying $\mathcal{F \circ \nabla}$ will leave you on the left side of $\mathcal{C + C}$. Because there are no morphisms between the two sides of $\mathcal{C + C}$, there can be no natural transformations between $\mathcal{F \circ \nabla}$ and $\mathrm{id}_{\mathcal{C + C}}$ in either direction, therefore $\mathcal F$ cannot be a left or right adjoint to $\nabla$.

However, this result surprises me a little because I would have assumed there to be some duality between the diagonal and codiagonal functors. Is my reasoning above wrong? If I'm not wrong, and there are no adjoints, then do there exist adjoint-like structures for $\nabla$ that are in some way dual to the adjoints of $\Delta$? Or am I just looking for symmetry where there is none?

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Your argument is correct, and you're basically looking for symmetry where there is none. 2-categorical duality, including duality questions involving adjoints, is a bit more complicated than the ordinary case.

The duality that sends the product to the coproduct is the opposite (2-) category: $C+C$ is the product in Cat*, as in any category. Now, an adjunction $f:x\leftrightarrow y:g$ in a 2-category gives rise to an adjunction $g^*:x^*\leftrightarrow y^*:f**$ in the opposite 2-category. The 2-morphisms have not been turned around, so we still have the 2-morphism corresponding to the unit of the original adjunction, $\eta^*: 1_{x^*}\to (gf)^*=f^*g^*$. Thus our adjunction has given rise to an adjunction in the opposite 2-category between the same objects, with left and right adjoints interchanged.

There's another dual of a 2-category, which is the one realized in Cat by sending $C$ to $C^*$ (note the difference from the duality above!) This one turns around the 2-arrows, so it also preserves adjunctions but switches the left and right adjoints.

So, what happens in your case? The previous paragraph says that the adjoints to $\Delta$ are still adjoints to $\Delta:C^*\times C^*\to C^*$, but with left and right switched: this is the familiar phenomenon that products and coproducts are interchanged between $C$ and $C^*$. The paragraph before that says that we get some adjunctions in Cat*, which is useless.

What we would need is adjoints to the diagonal in Cat*, which would become adjoints to the codiagonal back in Cat. For this you would need some sort of general principle guaranteeing you such adjoints in a sufficiently large class of 2-categories. But notice that, while Cat* certainly has products and coproducts, asking for these adjoints is asking that $C^*$ have products in some internal sense, as an object of Cat*, which is a totally different level of question. What we are seeing is that, just because an object of a 2-category has done limits and colimits in the internal sense, there's no reason at all to believe the same holds off the same object viewed in the opposite 2-category. The dualities are simply not acting at the same level.

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