Basic category theory: applying natural transformations

Question

So I'm starting to learn category theory, and I understand the definitions of functors and natural transformations - but I'm wondering, given a functor $F: \mathcal{A} \rightarrow \mathcal{B}$ and a natural transformation $\alpha: F \rightarrow G$ consisting of morphisms $\{\alpha_A: F(A) \rightarrow G(A) \}$, how is it that one could actually construct $G$?

Now clearly we can see where objects are mapped as we will have $G(A) = \alpha_A \circ F(A)$ for every object $A$. But my question is that, given a morphism $f: A \rightarrow A'$ in $\mathcal{A}$, how can we determine where this is mapped by $G$? Clearly it will be mapped to a morphism between $G(A)$ and $G(B)$ in $\mathcal{B}$, of which there is at least one. But which one?

Answer 1

You cannot in general construct $G$ from $\alpha$ and $F$. Rather, you should think about $G$ as something satisfying the conditions given by $\alpha$ and $F$.

I believe the following is an example of this ambiguity. Let $\mathcal{C}$ be the category with two objects $a, b$ and a unique nonidentity arrow $!: a\rightarrow b$. Consider the category $\mathcal{D}$ with four objects $x, y, z, w$ and nonidentity arrows

• $f: x\rightarrow y$

• $g_1: z\rightarrow w$

• $g_2: z\rightarrow w$

• $h_1: x\rightarrow z$

• $h_2: y\rightarrow w$

• $j: x\rightarrow w=h_2f$.

satisfying

$$g_1h_1=g_2h_1=j.$$ Now let $F:\mathcal{C}\rightarrow\mathcal{D}$ be the unique functor sending $a$ to $x$ and $b$ to $y$ (note that this must send $!$ to $f$), and let $G_1$ and $G_2$ be the two functors from $\mathcal{C}$ to $\mathcal{D}$ sending $a$ to $w$ and $b$ to $z$.

Then $\alpha=\{h_1, h_2\}$ is a natural transformation from $F$ to $G_1$ and from $F$ to $G_2$.

Answer 2

Actually, $\alpha$ does not determine $G$; the definition of natural transformation between two given functors $F,G : \mathcal{A} \to \mathcal{B}$ is as a collection of maps $\{\alpha_A:F(A)\to G(A)\}_{A\in Ob(\mathcal{A})}$ in $\mathcal{B}$ such that for all arrow $f:A\to B$ in $\mathcal{A}$ the diagram $$\require{AMScd} \begin{CD} F(A) @>{\alpha_A}>> G(A)\\ @V{F(f)}VV @VV{G(f)}V\\ F(B) @>>{\alpha_B}> G(B) \end{CD}$$ commutes, i.e. the equation $G(f)\circ \alpha_A = \alpha_B\circ F(f)$ holds in $\mathcal{B}$. For this definition to make sense, you need to know the functor $G$ beforehand (both on objects and arrows).

It is the same for sets and functions : an arrow $g:X\to Y$ does not determine $Y$ (nor $X$); you need to know $X$ and $Y$ before you can define $g$. Or you can take any category you want : in general, you define what are objects and then what are arrows between two objects.