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I've just run into the following categorical situation.

Let $\mathcal{A}_1,\mathcal{A}_2,\mathcal{C}, \mathcal{D}$ be categories, equipped with functors $F_1,F_2,G_1,G_2$ as follows:

$$\require{AMScd} \begin{CD} \mathcal{A}_1 @>{G_1}>> \mathcal{D}\\ @V{F_1}VV @AA{G_2}A \\ \mathcal{C} @<<{F_2}< \mathcal{A}_2 \end{CD}$$

I am interested in using the category $\mathcal{D}$ to define a "transformation" $F_1\to F_2$. Specifically, if $A_1,A_2$ are objects of $\mathcal{A}_1,\mathcal{A}_2$, I want to assign, to each morphism $G_1(A_1)\to G_2(A_2)$ in $\mathcal{D}$, a morphism $F_1(A_1)\to F_2(A_2)$ in $\mathcal{C}$.

This should be done in a functorial way, i.e. given morphisms $A_1'\to A_1$ and $A_2\to A_2'$, the morphisms associated to $G_1(A_1)\to G_2(A_2)$ and $G_1(A_1')\to G_2(A_2')$ should fit into a commutative diagram.

Is this a well-known construction, or does it arise in some natural way from other well-known constructions? I suspect that this should be a natural transformation between functors from some fibered category, but I have not managed to work out the details.


In my application, $\mathcal{A}_1, \mathcal{A}_2$, and $\mathcal{D}$ are thin, and $G_2$ is an embedding. I am also curious what conditions make this construction possible.

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  • $\begingroup$ If you want to use fibered categories pick $\mathcal{A}_1 \sqcup \mathcal{A}_2$ as the category lying above and defines a morphism between the categories with different bases ($\mathcal{C}$ and $\mathcal{D}$). Anyway, I don't think this will make computations easier. $\endgroup$ – user40276 Oct 16 '15 at 16:19
  • $\begingroup$ @user40276 I'm not sure I see how this works. Somehow the trick is to pick out the morphisms in $\mathcal{D}$ between objects in the images of $G_1$ and $G_2$. $\endgroup$ – Slade Oct 16 '15 at 18:25
  • $\begingroup$ It's pretty much unimaginable that this could work as you've stated it. What if the $F$s are isomorphisms but the $G$s are constant? At the very least, you need to settle for maps that are in the images of the $G$s and the $F$s, respectively. $\endgroup$ – Kevin Carlson Oct 16 '15 at 19:29
  • $\begingroup$ @KevinCarlson I have added some conditions in case they are useful, but I don't understand your objection. In the situation you describe, with $\mathcal{C}$ mapping to two objects $X,Y$ of $\mathcal{D}$, the goal would be to assign (in a compatible way) to each pair of objects $A,B$ in $\mathcal{C}$, and each morphism $X\to Y$, a morphism $A\to B$. $\endgroup$ – Slade Oct 16 '15 at 20:23
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    $\begingroup$ @KevinCarlson Sure, and there may not be a natural transformation between two functors with the same source and target. The actual definition still makes sense in generality—I'm not asking when this is possible, I'm asking if there is a good way to understand the concept. $\endgroup$ – Slade Oct 17 '15 at 0:34
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Not sure how useful this is to you, but you can describe this in terms of comma categories. Specifically, you are describing a functor $(G_1\downarrow G_2)\to(F_1\downarrow F_2)$ which commutes with the projection functors from the comma categories to $\mathcal{A}_1$ and $\mathcal{A}_2$. (Note that such a functor is entirely determined by what it does on objects, since the projections are jointly faithful.)

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  • $\begingroup$ Thanks! I'll have to check the details, but this looks like exactly what I was looking for. I had a feeling I was missing an important part of the language, and it looks like comma categories are it. $\endgroup$ – Slade Oct 17 '15 at 0:53
  • $\begingroup$ The details check out. It might not be exactly what I asked for, but it's what I meant to ask for, which is even better. $\endgroup$ – Slade Oct 17 '15 at 19:59

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