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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.

$\textbf{(2) Example:}$ A more interesting functor is the functor $-\times X_0:\textbf{Set}\to\textbf{Set},$ where $X_0$ is a fixed set, which sends each set $Q$ to the set $Q\times X_0,$ and sends each map $f:Q\to Q'$ to the map $f\times X_0:Q\times X_0\to Q'\times X_0:(q,x)\mapsto (f(q),x).$ [Thus $f\times X_0$ could be written as $f\times \mathrm{id}_{X_0}.$]

$\textbf{(3) Definition:}$ 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

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

$\textbf{(4) Exercise:}$ 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).

$\color{Red}{Questions:}$

In Exercise (4) above, I am having trouble understanding the notations $H(K,-)(L)=H(K,L)$ and $H(K,-)(f:L\to L')=H(\text{id}_K,f).$ First, is it a special case of (2) Example or (3) Definition above.

Second, does $H(K,-)(L)=H(K,L)$ mean: $(K,-)\xrightarrow{H} (K,L):(K,-)\mapsto H(K,L)$ for defining functor on objects, and $H(K,-)(f:L\to L')=H(\text{id}_K,f)$ means: $((K,L)\xrightarrow{f}(K,L'))\xrightarrow{H}(H(K,L)\xrightarrow{Hf}H(K,L'))=H(\text{id}_K,L)\xrightarrow{Hf}H(\text{id}_K,L')$

Thank you in advance

Seth
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    Think of the dash as a placeholder to be filled in. For a simpler example, if $f:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ is defined by $(x,y)\mapsto x^2-y^2$, then for fixed $a\in\mathbb{R}$ the function $f(a,-):\mathbb{R}\to\mathbb{R}$ is defined by $y\mapsto a^2-y^2$. – blargoner Sep 23 '23 at 05:32
  • @blargoner for functors defining on composition of morphisms, do I define it as follow: $H(K,-)(f\circ g:L\to L'\to L'')=H(\text{id}_K,f\circ g)=H(\text{id}_K,f)H(\text{id}_K,g)$? – Seth Sep 23 '23 at 06:23
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    Your first equality there is just from the definition of $H(K,-)$, but your second equality is part of what you must prove to show that $H(K,-)$ is actually a functor. You can't just say it's true by definition. – blargoner Sep 23 '23 at 13:53

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