TITLE: Uniqueness of completion of a metric space
QUESTION [3 upvotes]: I've been reading Aliprantis and Burkinshaw. This book gives a nicer construction of completion of metric space(in my opinion) avoiding any equivalence class or pseudometrics. It gives a simple isometric embedding of a metric space $(X,d)$ into  the set  of bounded functions on $X$ i.e $B(X)$ with the usual $sup$ norm metric by the following map $x\mapsto f_x:X\to\mathbb{R} $ where $f_x(a):=d(x,a)-d(a,\lambda)$ for some fixed $\lambda\in X$. One can see this is a bounded function and the map is an isometry and since $B(X)$ is complete we see that the closure of the image of $X$ in $B(X)$ under this map is complete as well. However I think the book pretty much "handwaves" the proof of uniqueness (see Page 46). The discussion begins by taking two completions $((Y_1,d_1),f_1)$ and $((Y_2,d_2),f_2)$ of $(X,d)$. Then $h:=f_2\circ f_1^{-1}:f_1(X)\to f_2(X)$ is a surjective isometry and since $f(X)$ is dense in $Y_1$ we extend $h$ to $h_*:Y_1\to Y_2$ (uniformly continuous). It still remains to show that $h_*$ is a surjection and an isometry. I do agree that this is intuitive but the proof is not as trivial as the book says "it is srtraightforward to show...". 
My proof for these two "not so straightforward" assertions are given below. Is there an obvious way to prove it that the authors can see but its not so apparent to me???
$Claim\; 1: h_*$ is onto.
To see this let $y_2\in Y_2$. Then wee choose a sequence $f_2(x_n)\to y_2$ since $f_2(X)$ is dense in $Y_2$. By isometries $f_1(x_n)$ is Cauchy so let $f_1(x_n)\to y_1\in Y_1$. So $h_*(y_1)=h_*(\lim f_1(x_n))=\lim h_*(f_1(x_n))=\lim h(f_1(x_n))=\lim f_2(x_n)=y_2$. This shows surjection.
$Claim\; 2: h_*$ is isometry.
To show that $d_1(y_1,y_1')=d_2(h_*(y_1),h_*(y_1'))$, we choose sequences $f_1(x_n)\to y_1$ and $f_1(x_n')\to y_1'$, then $d_2(h(f_1(x_n)),h(f_1(x_n')))\to d_2(h_*(y_1),h_*(y_1'))$. But $d_2(h(f_1(x_n)),h(f_1(x_n')))=d_2(f_2(x_n),f_2(x_n'))=d(x_n,x_n')=d_1(f_1(x_n),f_1(x_n'))\to d_1(y_1,y_1')$. Hence by Hausdorffness of $\mathbb{R}$ with usual topology $d_1(y_1,y_1')=d_2(h_*(y_1),h_*(y_1'))$.
All this  doesn't really seem that staraightforward. So is there some obvious way to show these assertions???

REPLY [1 votes]: Consider $k=f_1\circ f_2^{-1}\colon f_2(X)\longrightarrow f_1(X)$, which is a surjective isometry and, since $f(X)$ is dense in $Y_1$, we can extend it to $k^∗\colon Y_2\longrightarrow Y_1$. Both $k^*\circ h^*$ and $h^*\circ k^*$ are the identity in a dense subset of their domain. Therefore, since both are continuous functions, $k^\star\circ h^\star=\operatorname{id}$ and $h^\star\circ k^\star=\operatorname{id}$. In particular, $h^\star$ is surjective.
Your proof of the fact that $h^*$ is an isometry is fine.
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