Transpose of a diagram under an adjunction

Question

By a diagram of type $I$ ($I$ is also a category)in a category $C$, we mean a functor $F:I\longrightarrow C$.

Now, let’s assume we have an adjunction $F:C\longrightarrow D$ and $G:D\longrightarrow C$ such that $F\dashv G$.

With these data, how one can define the transpose of a diagram? Is there a kind of reverse operation?

I have some intuition that what the transpose of a diagram mean, but it is not so clear.

Answer 1

If $f:Fc\to d$ is an arrow in $D$, then what's usually called its transpose is the morphism $f^\flat:c\to Gd$ in $C$ obtained through the natural isomorphism coming from the adjunction:

$$D(c,Gd)\cong C(Fc,d)$$

Naturality of this bijection is characterized by saying that every left square as below commutes iff the right square commutes. This is like saying that the right square is the transpose of the left one and vice-versa.

While the transpose a generic diagram remains undefined (and I'm unsure there is such a concept), I believe this encodes all cases where you may want to tranpose through an adjoint. For instance, now we can take complicated diagrams in $D$ such as

and tranpose it to $C$, obtaining

All images were taken from Chapter 4 of Category Theory in Context.

Answer 2

As @Daniel Teixeira remarked in his answer, it only really makes sense to “transpose” morphisms of the form $Fc \to d$ where $c\in \mathcal{C}$ and $d\in \mathcal{D}$. In his example, we had a commutative square of the form: $\require{AMScd}$ \begin{CD} F(c) @>>> d\\ @V{}VV @VVV\\ F(c') @>>> d' \end{CD}

We can rephrase this a little bit — let $\mathcal{I} = 0\to 1$ be the category with two objects, and one morphism between them. Then such a commuative diagram on the left can be thought of as the composite of a diagram $A: \mathcal{I} \to \mathcal{C}$ and $F: \mathcal{C} \to \mathcal{D}$ (the left vertical arrow), a diagram $B:\mathcal{I} \to \mathcal{D}$ (the right vertical arrow) equipped with a natural transformation between them (yielding the horizontal arrows). That is, we can express the data of that diagram as a natural transformation $$ \phi: F\circ A \Rightarrow B.$$

Given such data, it is possible to "transpose" the diagram in the following way: using the unit $\eta$ of the adjunction we have a natural transformation

$$A = \text{id}_{\mathcal{C}}\circ A\overset{\eta\circ A}\Rightarrow G\circ F \circ A\overset{G\circ \phi}{\Rightarrow} G\circ B.$$

This gives us a function $[F\circ A, B] \to [A, G\circ B]$, and using the counit of the adjunction, you can check this is a (natural) bijection. This leads us to the following result.

Theorem: For a category $\mathcal{I}$, and functors $A: \mathcal{I} \to \mathcal{C}$, and $B: \mathcal{I} \to\mathcal{D}$, any diagram in $\mathcal{D}$ arising from a natural transformation $F\circ A\Rightarrow B$ can be transposed across the adjunction $F\dashv G$ to give a diagram in $\mathcal{C}$ assocated to a natural transformation $A\Rightarrow G\circ B$.

Note in particular that "triangles" of the form

fall under this result, if you just take $B: \mathcal{I} \to \mathcal{D}$ taking both objects to $d$.

I don't think you can phrase this theorem as an "if and only if" because there is no general notion of transposing an arbitrary diagram across an adjunction. However you could feasibly consider this result to define the types of diagrams that can be transposed.