The following question is taken from Arrows, Structures and Functors the categorical imperative by Arbib and Manes.
$\color{purple}{Background}$ $\color{purple}{Information}$
The category $\textbf{Met*}.$ Let $\textbf{Met*}$ be the category whose objects are metric spaces with base point $(X,d,x)$ (meaning $(X,d)$ is a metric space and $x$ "the base point" is an arbitrary element of $X$) and whose morphisms $f(X,d,x)\to (Y,e,y)$ are contractions $f:(X,d)\to (Y,e)$ such that $fx=y.$ It is clear that $\textbf{Met*}$ is a category since if $fx=y$ and $gy=z$ then $gfx=gy=z.$
Products in $\textbf{Met*}:$ Every family in $\textbf{Met*}$ has a product, although it need not be built on the cartesian product set! Given $(X_i,d_i,x_i)$ let $X$ be the subset of all $I-$tuples $(a_i)$ with each $a_i\in X_i$ such that the $I-$tuple of numbers $d_i(a_i,x_i)$ has an upper bound, and define $d((a_i),(b_i))=\text{Sup }\{d_i(a_i,b_i)\mid i\in I\}.$ Then $x=(x_i)$ is in $X.$ Since there exist, by definition of $X,$ numbers $s$ and $t$ such that $d_i(a_i,x_i)\leq t$ for all $i$ and $d_i(b_i,x_i)\leq s$ for all $i$ then $$d_i(a_i,b_i)\leq d_i(a_i,x_i)+d_i(x,b_i)=d_i(a_i,x_i)+d_i(b_i,x_i)\leq t+s$$ which proves that $d((a_i),(b_i))$ is a well -defined number. Let $p_i:(X,d_,x)\to (X_i,d_i,x_i)$ be the usual coordinate projections.
$\color{green}{Exercise:}$ Consider $\textbf{R}_+$ to be $\textbf{Met*}$ with the usual metric and $0$ as base point.
Show that every $(X,d,x)$ in $\textbf{Met*}$ is isomorphic to a subspace of a product (in $\textbf{Met*}$) of copies of $\textbf{R}_+$
[Hint: For any $y\in X, f_y:(X,d,x)\to \textbf{R}_+, z\mapsto d(y,z)-d(y,x)$ is a morphism in $\textbf{Met*}.$]
$\color{blue}{Question}$/$\color{blue}{difficulties:}$
From the hint, we have, $f_y(x)=d(y,x)-d(y,x)=0.$ Suppose $x_1\neq x_2\in X,$ then $e(f_y(x_1),f_y(x_2))=|f_y(x_1)-f_y(x_2)|=|(d(y,x_1)-d(y,x))-(d(y,x_2)-d(y,x))|=|d(y,x_1)-d(y,x_2)|\leq d(x_1,x_2)>0$
I am not sure what to do from here. My $\textbf{main difficulties}$ are several:
not knowing how to translate:
"subspace of a product (in $\textbf{Met*}$) of copies of $\textbf{R}_+$" into math notation, not clear on what the product of $\textbf{Met*}$ look like in terms of coordinates from the description I cited from the text. So if we let
$a=(a_i)_{i\in I}, b=(b_i)_{i\in I},$ then $d(a,b)=d((a_i), (b_i))=\text{Sup }\{d_i(a_i,b_i)\mid i\in I\}.$ I am not sure if this is correct.
I do know that I have to show the isomorphism is also an isometry. Lastly, I don't know how the hint is relevant to the question. Thank you in advance.
Let us embed $(X,d,x)$ (isometrically) into the $\textbf{Met*}$-product $(Y,d',{\bf0})$ of the family, indexed by $X$, of copies of $\textbf{R}_+=(\Bbb R,m,0)$ (where $m$ is the usual metric). This is Kuratowski's embedding.
$f:X\to Y$ is defined by
$$\forall y\in X\quad f(y)=f_y=(f_y(z))_{z\in X}$$ where $f_y$ was given by the hint: $$f_y(z)=d(y,z)-d(x,z).$$
$f(y)\in Y$ since $(|f_y(z)-0|)_{z\in X}$ is bounded (by $d(x,y)$).
$f$ is an isometry since $$\begin{align}d'(f(y_1),f(y_2))&=\sup_{z\in X}|f_{y_1}(z)-f_{y_2}(z)|\\&=\sup_{z\in X}|d(y_1,z)-d(y_2,z)|\\&=d(y_1,y_2)\end{align}$$ (because this upper bound is attained for instance when $z=y_1$).
This ends the proof.