The following question is taken from "Arrows, Structures and Functors the categorical imperative" by Arbib and Manes
$\color{Green}{Background:}$
$\textbf{(1a)}$ $\textbf{Definition:}$ 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.
$\textbf{(1b)}$ $\textit{The Category}$ $\textbf{Met}.$ Let $(X,d), (Y,e)$ be metric spaces. A $\textbf{contraction}$ from $(X,d)$ $\textit{to}$ $(Y,e)$ is a function $f:X\to Y$ satisfying $e(fx_1,fx_2)\leq d(x_1,x_2)$ It is clear that a contraction is Lipschitz (take $\lambda=1$). By adapting the proof for Lipschitz maps above it is clear that $g\cdot f$ is a contraction when $f$ and $g$ are and, of course, $\text{id}_X:(X,d)\to (X,d)$ is a contraction. We obtain the category $\textbf{Met}$ of metric spaces and contractions.
$\textbf{(2) Exercise:}$ Let $(X,d)$ be a metric space. Define a category $\textbf{K}$ as follows. An object is an element $x$ of $X.$ $\textbf{A}$ morphism $x\to y$ is $\lambda\in \textbf{R},$ such that $\lambda\geq d(x,y).$ If $\lambda:x\to y,$ $\mu:y\to z,$ $\mu\cdot \lambda:x\to z$ is defined to be $\mu+\lambda.$ Show that $\textbf{K}$ is a category.
Solution: To show that this is a category, first, the identity morphism: $0:x\to x$ is $\lambda=0=d(x,x),$ For composition of morphisms, let $\lambda:x\to y,$ $\mu:y\to z,$ be defined respectively as $\lambda\geq d(x,y)$ and $\mu\geq d(y,z)$, then $\lambda\cdot\mu:x\to z$ is defined aS $\lambda+\mu\geq d(x,y)+d(y,z)\geq d(x,z),$ by the triangle inequality. Hence for associativity of morphisms, suppose $p$ is another morphism $z\to w$ defined as $p\geq d(z,w).$ Then $p\cdot(\lambda\cdot \mu)$ is defined as $p+(\lambda+\mu)\geq d(x,y)+(d(y,z))+d(z,w))\geq d(x,y)+d(y,w)\geq d(x,w).$ Similarly for $(p\cdot\lambda)\cdot \mu,$ is define as $(p+\lambda)+\mu\geq (d(x,y)+d(y,z))+d(z,w)\geq d(x,z)+d(z,w)\geq d(x,w).$ Hence composition axiom for a category is satisfied.
$\textbf{(3) Exercise:}$ Let $f:(X,d)\to (Y,e)$ in $\textbf{Met,}$ and let $(X,d), (Y,e)$ be regarded as categories as in exercise (2). Show that $x\to fx, \lambda\mapsto \lambda$ makes $f$ a functor.
Solution: Let $K=(X,d),$ $K'=(Y,e)$ to be subcategories of $\textbf{Met},$ The objects in $K$ and $K'$ are respectively any elements $x\in X$ and $fx\in Y.$ Hence by $\textbf{(1a) Definition},$ we can define a functor $f$ from a category $K$ to a category $K'$ which maps $\text{Obj K}\to \text{Obj K'}:x\mapsto fx,$ and for each pair of objects $x,x'$ of $K,$ maps $K(x,x')\to K'(fx, fx'):\lambda\mapsto f\lambda.$ By (2) Exercise above, morphisms in $K$ and $K'$ are respectively $\lambda\geq d(x,x')$ and $\lambda\geq e(fx,fx').$ Hence we have $(\lambda\geq d(x,x')\xrightarrow{f(\lambda)} (\lambda\geq e(fx,fx')).$ For the case of functor for composition of morphisms. If $\lambda_1$ is another morphism in $K,$ with objects $x', x''\in K,$ then $\lambda_1\geq d(x', x'')$ and $\lambda_1\geq e(fx',fx'').$ are morphisms respectively in $K$ and $K'.$ Hence $(\lambda+\lambda_1\geq d(x,x')+d(x',x''))\xrightarrow{f(\lambda\cdot\lambda_1)} (\lambda+\lambda_1\geq e(fx,fx')+e(fx',fx''))=((\lambda\geq d(x,x')\xrightarrow{f(\lambda)} (\lambda\geq e(fx,fx')))((\lambda_1\geq d(x',x'')\xrightarrow{f(\lambda_1)} (\lambda_1\geq e(fx',fx''))).$
$\color{Red}{Questions:}$
For (3) Exercise solution above, I want to know if I show the functor for morphism and for composition of morphisms correctly. The exercise stated that for functor on morphism, it is $\lambda\mapsto \lambda.$ Should it not be written instead as $\lambda\geq d\xrightarrow{f(\lambda)} \lambda\geq e:\lambda\stackrel{f(\lambda)}{\mapsto} \lambda?$
Thank you in advance.