Showing functor for composition of morphism two variable functor with one fixed variable?

Question

The following question is taken from "Arrows, Structures and Functors the categorical imperative" by Arbib and Manes

$\color{Green}{Background:}$

$\textbf{(1)}$ $\textbf{Definition:}$ A functor $H$ from a category $\textbf{K}$ to a category $\textbf{L}$ is a function which maps $\text{Obj}\textbf{(K)}\to \text{Obj}\textbf{(L)}:A\mapsto HA,$ and which for each pair $A,B$ of objects $\textbf{K}$ maps $\textbf{K}(A,B)\to \textbf{L}(HA, HB):f\mapsto Hf,$ while satisfying the two conditions:

$$H(\text{id}_A)=\text{id}_{HA}\quad\text{ for every }A\in\text{Obj}\textbf{(K)}$$ $$H(g\cdot f)=Hg\cdot Hf \quad\text{ whenever }g\cdot f\text{ is defined in }\textbf{K}.$$

We say that $H$ is an $\textbf{isomorphism}$ if $A\mapsto HA$ and each $\textbf{K}(A,B)\to \textbf{L}(HA, HB)$ are bijections.

$\textbf{(2)}$ $\textbf{Definition:}$ Given two categories $\textbf{K}$ and $\textbf{L},$ we define their $\textbf{product}$ $\textbf{K}\times \textbf{L}$ to be the category whose objects are ordered pairs $(K,L)$ of objects $K$ from $\textbf{K}$ and $L$ from $\textbf{L},$ and for which morphisms

$$(K,L)\to (K',L')$$

are just pairs $(f,g)$ with $f\in \textbf{K}(K,K')$ and $g\in \textbf{L}(L,L'),$ while

$$\mathrm{id}_{(K,L)}=(\mathrm{id}_K,\mathrm{id}_L)\text{ and }(f',g')\cdot(f,g)=(f'\cdot f,g'\cdot g).$$

To tie all these concepts together we note the following:

$\textbf{(3)}\textbf{ Observation:}$ The map which assigns to each pair of objects $K,K'$ in the category $\textbf{K}$ the set $\textbf{K}(K,K')$ of morphisms from $K$ to $K'$ becomes a functor

$$\mathrm{hom}:\textbf{K}^\mathrm{op}\times \textbf{K}\to \textbf{Set}:(K,K')\mapsto \textbf{K}(K,K')$$

when we make the morphism assignment

$$(f:K_1 --<{K_1}^{'}, g:K_2\to {K_2}^{'})\mapsto \textbf{K}(K_1,K_2)\xrightarrow{g\cdot(-)\cdot f}\textbf{K}({K_1}^{'},{K_2}^{'}).$$

($\textbf{hom}$ is sometimes referred to as the $\textbf{external representation functor,}$ and is often written $\textbf{K}(\cdot,\cdot).$)

$\textbf{(4) Exercise:}$ If $H:\textbf{K}\times \textbf{L}\to \textbf{N}$ is a functor of two variables and if $K\in \textbf{K}$ is a fixed object, then show that $H(K,-):\textbf{L}\to \textbf{N}$ defined by $H(K,-)(L)=H(K,L), H(K,-)(f:L\to L')=H(\text{id}_K,f)$ is a functor $\textbf{L}\to \textbf{N}$ of (one variable).

$\color{Red}{Questions:}$

I have asked about the above exercise here and here. I posted twice about trying to understand the functor notation used in the exercise. I understand that $H(K,-)(L)=H(K,L),$ means $(K,-)\xrightarrow{H} (K,L):(K,-)\mapsto H(K,L)$ for defining functor on objects.

As for $H(K,-)(f:L\to L')=H(\text{id}_K,f)$ is a functor for defining on morphism. In terms of mapping notation, it translates to $((K,L)\xrightarrow{f}(K,L'))\xrightarrow{H}(H(K,L)\xrightarrow{Hf}H(K,L'))=H(\text{id}_K,L)\xrightarrow{Hf}H(\text{id}_K,L').$ I am not sure if that is the correct interpretation. If so, for the case of defining functor for composition of morphism, would it mean: $H(K,-)(f\circ g:L\to L'\to L'')=H(\mathrm{id}_K,f\circ g)=H(\mathrm{id}_K,f)\cdot H(\mathrm{id}_K, g)?$ Then to show the correctness of this equational identity.

Might it be the following: $H(K,-)(f\circ g:L\to L'\to L'')=[((K,L)\xrightarrow{f}(K,L'))\xrightarrow{H}(H(K,L)\xrightarrow{Hf}H(K,L'))] \cdot [((K,L)\xrightarrow{g}(K,L'))\xrightarrow{H}(H(K,L')\xrightarrow{Hg}H(K,L''))]=[H(\text{id}_K,L)\xrightarrow{Hf}H(\text{id}_K,L')]\cdot [H(\text{id}_K,L')\xrightarrow{Hg}H(\text{id}_K,L'')]=H(\mathrm{id}_K,f)\cdot H(\mathrm{id}_K, g)=H(\mathrm{id}_K,f\circ g).$ I am not sure if this is correct.

Answer 1

Let's denote $H(K,-)$ by $G$. So, $G(X)=H(K,X)$ on objects and $G(f)=H(1,f):G(X)\to G(X')$ for $f:X\to X'$. I interchange "$1$" and "$\mathrm{id}_K$" at my convenience. They represent the same thing.

$$(K,-)\overset{H}{\longrightarrow}(K,L):(K,-)\mapsto H(K,L)$$

In my opinion, this means nothing. It looks like an attempt to use domain-codomain-mapping notation for a function, e.g. $f:\Bbb R\to\Bbb R:\,x\mapsto x^2$ but $(K,-)$ is neither a 'domain' nor an object of a domain and $H(K,-)=(K,L)$ is (sorry) just nonsense.

The 'mapping notation' you use; it's not correct and it also has great potential to be confusing, I would recommend you avoid using it until you're confident about what goes where.

$$((K,L)\xrightarrow{f}(K,L'))\xrightarrow{H}(H(K,L)\xrightarrow{Hf}H(K,L'))=H(\text{id}_K,L)\xrightarrow{Hf}H(\text{id}_K,L')$$

$f$ is not supposed to be an arrow $(K,L)\to (K,L')$; in making sense of the functor $G=H(K,-)$ we would only consider arrows in $\mathbf{L}$ not $\mathbf{K}\times\mathbf{L}$, so I'd expect to see $f:L\to L'$. Then we have a bit of an error-carried-forward; I don't want to see $Hf:H(K,L)\to H(K,L')$, I want to see $G(f)=H(1,f):H(K,L)\to H(K,L')$.

Then $H(\mathrm{id}_K,L)$ etc. is just ... meaningless, $\mathrm{id}_K$ is an arrow of $\mathbf{K}$ and almost never denotes an object of $\mathbf{K}$, and $L$ is an object not an arrow. $H(\text{arrow},\text{arrow})$ makes sense and $H(\text{object},\text{object})$ makes sense - $H(\text{arrow},\text{object})$ does not make sense, sadly.

Problem: Is $G$ a functor?

Is $G(f\circ g)=G(f)\circ G(g)$ for $L\overset{g}{\to}L'\overset{f}{\to}L''$ in $\mathbf{L}$? In other words, is $H(1,f\circ g)=H(1,f)\circ H(1,g)$?

Also, is $G(\mathrm{id}_L)=\mathrm{id}_{G(L)}$ for all $L$? That is, is $H(\mathrm{id}_K,\mathrm{id}_L)=\mathrm{id}_{H(K,L)}$ for all $L$?

However, your 'proof' for this is problematic because it is using the erroneous/confusing/meaningless 'mapping notation'.

So, how do we show $H(1,f\circ g)=H(1,f)\circ H(1,g)$? Here's my hint: try to use the fact $H$ itself is a functor - the right hand side is a composition of type $H(\text{arrow})\circ H(\text{arrow})$, so you should be able to use (one of) the defining feature(s) of a functor to resolve this expression.