I'm having trouble with an exercise from Rotman's Homological Algebra. It has to do with what Wikipedia calls "horizontal composition" of natural transformations. Namely, given $F, G:\mathcal{A}\to\mathcal{B}$ and $F^\prime, G^\prime:\mathcal{B}\to\mathcal{C}$ covariant functors and $\sigma:F\to G$ and $\tau:F^\prime\to G^\prime$ natural transformations, the goal is to show that there is a composite natural transformation $\tau\sigma: F^\prime F\to G^\prime G$.
There should be a "natural choice" for the morphism $(\tau\sigma)_A$ associated to any $A\in\operatorname{obj}\mathcal{A}$, and this is where I'm confused. The problem says to define $$(\tau\sigma)_A=\tau_{FA}\sigma_A : F^\prime F(A)\to G^\prime G(A)$$ but this doesn't make sense to me since $\tau_{FA}\in\operatorname{Hom}(F^\prime F(A),G^\prime F(A))$ and $\sigma_A\in\operatorname{Hom}(FA,GA)$ (right?). It's not clear to me why this is the right composition, or that it even is a composition.
If I had to guess, I would say $(\tau\sigma)_A:=\tau_{GA}F^\prime(\sigma_A)$ because this is the only thing I can come up with that actually is a morphism $F^\prime F(A)\to G^\prime G(A)$.
Can anyone clear up my confusion?
Your guess is correct, but besides $\tau_{GA}\circ F'(\sigma_A)$, there is another natural choice, namely: $$(\tau\sigma)_A=G'(\sigma_A)\circ \tau_{FA}\,,$$ and the point is that they are equal.
To be honest, I have no idea where the $\tau_{FA}\sigma_A$ could come from.. Might be a typo..
Note also that if we define natural transformations as functors $\mathcal A\to\mathcal B^\to$, where $\mathcal B^\to$ denotes the arrow category of $\mathcal B$ whose objects are the arrows of $\mathcal B$ and whose morphisms are commutative squares in $\mathcal B$, then we arrive to