Uniform continous extension in a completion of metric space

Question

Maybe this question has been done before, but english is not my first language, so I could not thoroughly search for the question (Due to lack of notation knowledge)

I am trying to prove this proposition:

Let (M',d') be the completion of the metric space (M,d) and i: M to M' the inyective isometry that makes i(M) dense in M'. Let (N,d'') be a metric space and f: M to N, an uniformy continuous function, then there exists an uniformly continous function f': M' to N, such that f'o i = f.

Ok.. so what i thought was this: You can define f' in i(M) like so: f'(x)=f(i^(-1)(x)). My problem is defining it for i(M) compliment. I tought of using the density of i(M), and the fact that uniform continuous functions preserve Cauchy property in sequences, to construct a sequence xn in i(M) for every point x in i(M) compliment. Such that xn tends to x, and defining f'(x)= lim f(i^(-1)(xn)). But since N is not necesarily complete, this sequence doesnt necesarily converge. Cannot choose a fixed point in i(M) because all metric spaces are Haussdorf and it will ruin my unif. continuiy.

Any suggestions?

P.S. Sorry for the sloppiness of the question, hopes the point gets across though.

Answer 1

Your idea of extending $f$ by using the density of $i(M)$ is the right thing to do. Also, you are right when you mention that such approach only works when $N$ is complete. In fact, your proposition fails when $N$ is not complete. To see this, consider $M=N=(0,1)$ with the metric induced by the absolute value on $\mathbb{R}$. Here $M'=[0,1]$ and $i$ is the inclusion of $(0,1)$ into $[0,1]$.

Claim: There is no continuous function $f':[0,1]\to(0,1)$ extending the function $f:(0,1)\to(0,1)$ defined by $f(x)=x.$

Proof of the claim: Suppose for a moment that such $f'$ exists. Then we have the following: $$(0,1)=f((0,1))\subset f'([0,1]).$$ Hence $f'([0,1])=(0,1)$. Since the image of a compact set by a continuous function is a compact set, we have that $f'([0,1])=(0,1)$ is compact which is absurd. Therefore such $f'$ doesn't exist.

PS: Since you are already studying topology it's time for you to learn how to use LaTeX.