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From Categories for the Working Mathematician pg. 43:

Theorem 1. The collection of all natural transformations is the set of arrows of two different categories under two different operations of composition, $\cdot$ and $\circ$, which satisfy the interchange law (5).

Question 1: What are these "two different categories"? The author never specifies.

Moreover, any arrow (transformation) which is an identity for the composition $\circ$ is also an identity for composition $\cdot$.

Question 2: What is an example of an identity for the composition $\circ$ also serving as an identity for the composition $\cdot$? And is this just an example where for $\tau, \sigma$ natural transformations then

$$ \tau \cdot I = \tau = I \cdot \tau \iff \tau \circ I = \tau = I \circ \tau $$

holds?

Note the objects for the horizontal composition $\circ$ are the categories, for the vertical composition, the functors.

Question 3: This doesn't make sense to me. Aren't the objects for horizontal composition the horizontal morphisms between categories (whose objects are categories)?

For example, if $C$ and $D$ are categories, then it seems to me the author is (nonsensically) saying that $C \circ D$ makes sense.

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The horizontal category is:

  • Objects are categories
  • Morphisms are natural transformations
  • The product of morphisms is horizontal composition $\circ$
  • The identity for an object $C$ is $1_{1_{\mathcal{C}}}$

The vertical categtory is

  • Objects are functors
  • Morphisms are natural trasnformations
  • The product of morphisms is vertical composition $\cdot$
  • The identity for an obect $F$ is $1_F$

And note that in the vertical category, if $C$ is a category, the identity for the object $1_\mathcal{C}$ is $1_{1_{\mathcal{C}}}$


Incidentally, a category is determined (up to isomorphism) by its set of morphisms and composition law; there are even "arrow only" axiomatizations of categories (basically, you replace the notion of "object" with that of "identity arrow"). So, the text of theorem 1 actually does specify what the two categories are.

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Ad Q1: Those two categories are

  • category "Dot" where objects are functors, morphisms natural transformations and the composition of morphisms given by the $\cdot$ operation

  • category "Circ" with categories as objects and morphisms natural transformations between functors between them, the composition of morphisms is given by $\circ$ operation.

If you imagine "building block" of a general category to be the image $ \stackrel{x}\cdot \rightarrow \stackrel{y}\cdot$, then the building block for both categories are of the form $$\require{AMScd} \begin{CD} A @>F>> B \\ @V{\tau}VV @V{\tau}VV\\ A @>G>> B \end{CD}$$ ($\tau$ is a natural transformation from functor $F$ to $G$) and in the "Dot" category, you compose such images under (so two such images can be composed iff the initial/target functors coincide), while in the "Circ" category you compose next to (so two such images can be composed iff the initial/target categories coincide) - it actually implies what the respective objects are (those entities that have to coincide).

Ad Q2:

Note now, that if you compose two morphisms in "Circ", i.e. \begin{CD} A @>F_1>> B @>F_2>> C\\ @V{\tau}VV @V{\tau}V{\sigma}V @V{\sigma}VV\\ A @>G_1>> B @>G_2>> C \end{CD}

in such a way that $\sigma \circ \tau = \sigma$ then, since $\sigma: F_2 \rightarrow G_2$ and $\sigma \circ \tau: F_2F_1\rightarrow G_2G_1$, $F_1 = G_1 = 1_B$ and $A=B$.

If there is a morphism $\rho$ (natural transformation) from $1_B \rightarrow H$ of endofunctors on $B$ and you want to see, that $\tau$ behaves as identity, represent the $\cdot$ composition

\begin{CD} B @>1_B>> B \\ @V{\tau}VV @V{\tau}VV\\ B @>1_B>> B \\ @V{\rho}VV @V{\rho}VV\\ B @>H>> B \end{CD}

as

\begin{CD} B @>1_B>> B @>H>> B\\ @V{\tau}VV @V{\tau}V{\rho}V @V{\rho}VV\\ B @>1_B>> B @>H>> B \end{CD}

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  • $\begingroup$ I would label the whole square $\tau$, rather than the vertical arrows. (I would have used equality for the vertical arrows) $\endgroup$ – Hurkyl Feb 8 '17 at 17:42
  • $\begingroup$ I would use just one arrow in the middle, but the amscd cannot do that. And in this way, I hope, it still is clear. Well if not, feel free to edit it! $\endgroup$ – pepa.dvorak Feb 8 '17 at 17:48

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