Can natural transformations be viewed as functors between images of functors?

Let $$C$$ and $$D$$ be categories, and let $$F,G:C\rightarrow D$$ be functors. Then a natural transformation $$\tau$$ from $$F$$ to $$G$$ is a family of morphisms $$\{\tau_x\}_{x\in C}$$ where for each $$x\in X$$, $$\tau_x$$ is a morphism between $$F(x)$$ and $$G(x)$$. And this family has to satisfy a certain commutative diagram. But I'm wondering whether natural transformations can be viewed in a different way.

An image of a functor isn't always a subcategory, but suppose that $$F(C)$$ and $$G(C)$$ are subcategories of $$D$$. Then my question is, can $$\tau$$ be viewed as a functor between the categories $$F(C)$$ and $$G(C)$$? It sends $$F(x)$$ to $$G(x)$$ for any object $$x\in C$$, and it sends $$F(f)$$ to $$\tau_y\circ F(f)$$ for any morphism $$f:x\rightarrow y$$ in $$C$$.

Or does none of this make sense?

• $\tau_y\circ F(f)$ is a morphism from $F(x)\to G(y)$, so as an operation on morphisms it is certainly not doing the work of a functor. – Malice Vidrine Feb 8 at 4:45
• well, I see where you are going with your question, but I think you are going a little too far. In my opinion the best understanding for natural transformations is that they are morphisms in the category of functors. So no, they are sadly not functors, but they are morphisms between those. However, if you want you can of course interpret any morphism as a functor, however, you are gaining little information thanks to that – Enkidu Feb 8 at 7:26

1 Answer

The functor you have proposed is not necessarily well-defined; consider when there exist $$x,x' \in C$$ with $$Fx=Fx'$$ but $$Gx \neq Gx'$$.

I encourage you to research the notion of comma category. Very briefly, given functors $$F: \mathscr{A} \rightarrow \mathscr{C}$$ and $$G:\mathscr{B} \rightarrow \mathscr{C}$$, the objects of the comma category $$(F \Rightarrow G)$$ are morphisms $$(A,B,f:FA \to GB)$$ in $$\mathscr{C}$$ and the morphisms of the comma category $$(F \Rightarrow G)$$ are pairs $$(j:A \rightarrow A',k:B \rightarrow B') \in \mathscr{A}(A,A') \times \mathscr{B}(B,B')$$ such that the obvious square commutes. Notice that the objects of $$(F \Rightarrow G)$$ keep track of the objects $$A \in \mathscr{A}, B \in \mathscr{B}$$, and do not work only with the objects in the image, and that an arbitrary $$f:FA \rightarrow GB$$ need not be in the image of $$F$$.

Suppose $$F,G:\mathscr{A} \rightarrow \mathscr{B}$$. Let $$\mathfrak{N}$$ denote the collection of natural transformations $$F \rightarrow G$$. Given a natural transformation $$\alpha: F \rightarrow G$$, we can form the functor $$T_{\alpha}:A \rightarrow (F \Rightarrow G)$$ with $$TA=(A,A,\alpha_A)$$ and $$Tf=(f,f)$$. Let $$\mathfrak{R}$$ denote the collection of functors $$T:C \rightarrow (F \Rightarrow G)$$ with the property that $$\forall A \in \mathscr{A}$$ $$TA=(A,A,f_A)$$ for some $$f_A:FA \to GA$$ and $$\forall h \in \mathscr{A}(A,A') \ \ Th=(h,h)$$. Given such a functor $$T$$, we can form the collection of morphisms $$\alpha_T = \left( (\alpha_T)_A \right)_{A \in \mathscr{A}} = \left( f_A \right) _{A \in \mathscr{A}}$$.

https://ncatlab.org/nlab/show/comma+category#examples asserts that this $$\alpha_T$$ is a natural transformation, and therefore that this is a one-to-one correspondence between $$\mathfrak{N}$$ and $$\mathfrak{R}$$. I am not sure about this as the naturality condition for $$\alpha_T$$ seems to be identical to stating that $$\forall A,A' \in \mathscr{A}$$ $$(F \Rightarrow G)(TA,TA') = \mathscr{A}(A,A') \times \mathscr{A}(A,A')$$, which is not necessarily the case.

• You seem to have missed the part on the nlab page that says that a functor $T : C\to F/G$ gives a natural transformation, if both projections $F/G\to C$ are left inverses for $T$. – jgon Feb 23 at 0:40