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?

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    $\begingroup$ Given F and $\alpha$, there is no way to determine G. I would say that the question does not really make sense at all, in fact: you cannot have $\alpha$ if you do not have G already. And if you do have $\alpha$, then of course you can determine G: it is the codomain of $\alpha$! $\endgroup$ – Mariano Suárez-Álvarez Jun 3 '17 at 18:41
  • $\begingroup$ @MarianoSuárez-Álvarez I think this question does make sense, it's just based on an incorrect expectation. $\endgroup$ – Noah Schweber Jun 3 '17 at 18:49
  • $\begingroup$ Interesting. That seems counterintuitive when compared to the kinds of morphisms I've previously encountered, but of course there's no reason one shouldn't be able to do that. @Noah: yes, it must be exactly that - I am very new to this, and self-studying. $\endgroup$ – Chris Jun 3 '17 at 18:49
  • $\begingroup$ What kind of morphisms have you previously encountered? Essentially all morphisms (in modern language) determine their codomain. $\endgroup$ – Mariano Suárez-Álvarez Jun 3 '17 at 19:36
  • $\begingroup$ I suppose that makes sense actually. I guess in the category of continous functions and topological spaces, I don't expect the image of a function to determine the space. Neither does the 'image' of a natural transformation determine the functor it maps to. This is okay, although I think perhaps I'm having a little trouble wrapping my head around the motivation for the definition - it makes some sense, but I don't yet see that why it is the definition that makes sense. $\endgroup$ – Chris Jun 3 '17 at 19:55

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

  • $f: x\rightarrow y$

  • $g_1: z\rightarrow w$

  • $g_2: z\rightarrow w$

  • $h_1: x\rightarrow z$

  • $h_2: y\rightarrow w$

  • $j: x\rightarrow w=h_2f$.


$$g_1h_1=g_2h_1=j.$$ 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 $G_2$.

  • $\begingroup$ In fact we can easily construe the $\mathcal{D}$ in this counterexample as a (non-full) subcategory of Set. $\endgroup$ – Noah Schweber Jun 3 '17 at 18:47
  • $\begingroup$ Okay, that's great - thanks for the example. That was exactly the kind of situation I was wondering about. Although I guess in this case $G_1$ and $G_2$ are naturally isomorphic - is it the case then that a natural transformation and an input would determine its image up to natural isomorphism? EDIT: Gave this some thought, and since a natural transformation and an input determines the image of the objects, this is trivially true. Simply take the components of the natural isomorphism to be the identity map in every case. $\endgroup$ – Chris Jun 3 '17 at 18:58

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.


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