# Questions regarding a particular category with functors

I have two questions regarding categories and natural transformations.

The first question is as follows. Say we have categories $A$, $B_1$ and $B_2$, with functors $F: A \to B_1$, $G: B_1 \to B_2$ as well as functors $R_A$, $R_{B_1}$ and $R_{B_2}$ from the above mentionned categories to $Sets$. Assume now that we have natural transformations $\eta_1$ between $R_A$ and $R_{B_1} \circ F$, and $\eta_2$ between $R_{B_1}$ and $R_{B_2} \circ G$. I want to show that this implies that we have a natural transformation between $R_A$ and $R_{B_2} \circ G \circ F$ and that $\eta_2 \circ \eta_1$ works as such. I wrote down the definition and it seems to work, but can anyone confirm ?

The second question assumes that the result above is valid. Let's define a category in which objects are a set of two categories $A_1$, $B_1$ with functors $F: A_1 \to B_1$, $R_{A_1}: A_1 \to Sets$ and $R_{B_1}: B_1 \to Sets$, such that there exists a natural transformation between $R_{A_1}$ and $R_{B_1} \circ F$. Morphisms of this category are functors $U: A_1 \to A_2$, $V: B_1 \to B_2$ with natural transformations between $R_{A_1}$ and $R_{A_2} \circ U$ and between $R_{B_1}$ and $R_{B_2} \circ V$ such that the whole diagram of natural transformations commutes. I'd like to check if this category is a topos, but I have no idea of the methodology I should use. How should I procede ?

## 1 Answer

Your first answer doesn't work. $\eta_1$ is a transformation between functors $A \to \mathbf{Set}$, and $\eta_2$ is a transformation between functors $B_1 \to \mathbf{Set}$. Because $B_1 \neq \mathbf{Set}$, the horizontal composite $\eta_2 \circ \eta_1$ is not defined.

Let's make a diagram

$$\begin{matrix} A &-&-& - & - & \xrightarrow{R_A} & \mathbf{Set} \\ || & & & \Downarrow \eta_1 & & & || \\ A & \xrightarrow{F} & B_1 & - & - & \xrightarrow{R_{B_1}} & \mathbf{Set} \\ || & & || & & \Downarrow \eta_2 & & || \\ A &\xrightarrow{F} & B_1 &\xrightarrow{G}& B_2 &\xrightarrow{R_{B_2}} & \mathbf{Set} \end{matrix}$$

The top rectangle, for example, depicts $R_A$ and $R_{B_1} \circ F$ as functors $A \to \mathbf{Set}$, and that $\eta_1$ is a natural transformation between them. (I prefer my pictures orthogonally aligned, rather than the usual way people draw pictures. This has the advantage of being renderable in TeX!)

Having the picture makes it clear how the composition works. The bottom half of the picture is the horizontal composite $\eta_2 \circ F$. Or if you prefer, is $\eta_2 \circ \mathbf{1}_F$.

The natural transformation you're looking for is the vertical composite of the two halves:

$$(\eta_2 \circ F) \cdot \eta_1$$

• Thanks for your answer. I had another way of drawing the diagram, but yours is definitely better (kudos for doing it btw). I think I was sloppy when I spoke about $\eta_2 \circ \eta_1$, as what I meant was the natural transformation which, for each object $X$ of $A$ associates the morphism $\eta_{2_{F(X)}} \circ \eta_{1_X}$ (I guess I wrote $\eta_2 \circ \eta_1$ thinking about the functions between each sets). This would be the same as your answer $(\eta_2 \circ \mathbf{1}_F) \cdot \eta_1$, wouldn't it ? – OliverX1 Oct 21 '13 at 22:03
• @user: Working in the way I have, the easiest way to describe the pointwise action is to represent a point of $A$ as a functor $p : \mathbf{1} \to A$, where $\mathbf{1}$ is the category with one object, and just the identity arrow. Add this to the left of my diagram. The top half of the diagram is now $\eta_1\circ p$, and the bottom half is $\eta_2 \circ F \circ p$, so yes, we do get the arrow ${\eta_2}_{F(X)} \cdot {\eta_1}_{X}$. Aside: it's very convenient to use the same symbol (usually $\cdot$) for composition within a category as you do for vertical composition of natural transformations. – Hurkyl Oct 21 '13 at 22:34