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

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$\textbf{(1)}$ $\textbf{Definition:}$ 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.

$\textbf{(2) Exercise:}$ 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).

$H(K,-)(L)=H(K,L),$ is functor defined on objects, while $H(K,-)(f:L\to L')=H(\text{id}_K,f)$ is functor defined on morphisms.

For the case of the identity functor, we have:

$H(K,-)(i:L\to L)=H(\text{id}_K,i)$ and for defining functor for the case of composition of morphisms, we have:

$H(K,-)(f\cdot g:L\to L'\to L'')=H(K,-)(f:L\to L')(g:L'\to L'')=H(\text{id}_K,f)H(\text{id}_K,g)$

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What I would like to know is if how I describe the functor for composition of morphisms is correct? If not, can someone please give me the proper corrections please.

Thank you in advance.

Seth
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    You haven't proven anything here. You haven't proven $H(K,-)$ actually defines a functor. $H(K,-)(f\cdot g)=H(K,-)(f)(g)=H(1,f)H(1,g)$ isn't a proof, the quantity in the middle doesn't really make sense. It's just pushing symbols around – FShrike Sep 11 '23 at 22:30

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