Re: Function composition of natural transformations?

[Jpdonaly@aol.com]

For function composition, I just use the standard small circle \circ. So the
function composite of natural transformations \tau and \sigma (if it exists)
is \tau\circ\sigma. It is advisable to give up subscripting as a way of
denoting values of (fully extended) natural transformations: The value of \tau at
morphism a is just \tau(a). I would not use juxtaposition or any other generic
means (e.g., a centered dot) of denoting composition in a general category for
function composition or, for that matter, for any other composition which
already has a specified composition symbol, but I do denote pointwise ("vertical")
composition generically. Here is an example of how this goes---a line proof of
the interchange law for function and pointwise composition:

{\noindent\bf Proposition (Interchange Law):} When $\nu\mu\circ\tau\sigma$ is
defined for natural transformations $\nu$, $\mu$, $\tau$ and $\sigma$, then
so is $(\nu\circ\tau)\cdot(\mu\circ\sigma)$, and
$$\nu\mu\circ\tau\sigma=(\nu\circ\tau)\cdot(\mu\circ\sigma).$$
\medskip

{\noindent\bf Proof:} The void cases are trivial. Assume that
$\nu\mu\circ\tau\sigma$ is defined. Then surely $\nu\circ\tau$ and $\mu\circ\sigma$ are
defined, and
$$\dom(\nu\circ\tau)=\dom\nu\circ\dom\tau=\cod\mu\circ\cod\sigma=\cod(\mu\circ
\sigma),$$
so $\nu\circ\tau$ composes pointwise with $\mu\circ\sigma$. Calculate as
follows at an $a$ in the common domain category of both sides of the interchange
formula:
$$\eqalign{[(\nu\cdot\mu)\circ(\tau\cdot\sigma)](a)=&\nu[\tau(\cod
a)\cdot\sigma(a)]\cdot\mu(\dom[\tau(\cod a)\cdot\sigma(a)])\cr
&\cr
=&\nu(\tau(\cod a))\cdot(\dom\nu)[\sigma(a)]\cdot\mu[\dom\sigma(a)]\cr
&\cr
=&(\nu\circ\tau)(\cod a)\cdot(\cod\mu)[\sigma(a)]\cdot\mu[\dom\sigma(a)]\cr
&\cr
=&(\nu\circ\tau)(\cod a)\cdot\mu(\sigma(a))\cr
&\cr
=&(\nu\circ\tau)(\cod a)\cdot(\mu\circ\sigma)(a)\cr
&\cr
=&[(\nu\circ\tau)\cdot(\mu\circ\sigma)](a).\cr}$$
So the two sides of the interchange equation have the same intertwining
function.
Checking domain functors,
$$\eqalign{\dom(\nu\mu\circ\tau\sigma)&=\dom\nu\mu\circ\dom\tau\sigma\cr
&=\dom\mu\circ\dom\sigma\cr
&=\dom(\mu\circ\sigma)\cr
&=\dom(\nu\circ\tau)(\mu\circ\sigma);\cr}$$
similarly, $\cod(\nu\mu\circ\tau\sigma)=\cod(\nu\circ\tau)(\mu\circ\sigma)$.
Thus the two natural transformations are equal.

In this, \dom and \cod are defined by
\def\dom {\hbox{\rm dom }}
\def\cod {\hbox{\rm cod }}
and respectively represent the domain and the codomain function on the
implicit category. The proof uses the following formulas for pointwise composition
in terms of fully extended natural transformations (i.e.,  in terms of their
intertwining functions \pi and \tau):
(\pi\cdot\tau)(a)=\pi(a)\cdot\tau(\dom a)=\pi(\cod a)\cdot\tau(a)
which I can't help mentioning as an aside shows that evaluation of fully
extended natural transformations at a morphism intertwines evaluation at its
domain object with evaluation at its codomain object. (And, incidentally, codomains
are on the left in my notations, domains on the right.)

If I haven't explained something necessary here, I hope that you can
nevertheless see that the above line proof represents a moderately massive amount of
diagram drawing and chasing and would fit convincingly on the page of a
textbook. I hope that this addresses your request. The only examples which I know are
all in my personal set of notes which I set up PCTex32 over the last dozen
years and which come out at about 200 pages. This is probably a little too much
to drop on you all at once. I am, however, anxious to answer any further
questions which you may have.