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

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


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.


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.


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}


\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}

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.