This problem is from the Rutgers 1993 Sample Qual
Given metric spaces $X,Y$ and a surjective continuous function $f: X \rightarrow Y$ such that $d_X(x,y) \le d_Y(f(x), f(y)) $ prove or find a counter example to the following:
If $Y$ is complete then $X$ is complete.
I have found several examples on Math.Stackexchange of proving that if $X$ is complete then $Y$ is necessarily complete but this other direction seems to be a dead end.
One idea of a counterexample is let $X = [-1, 1), Y = [0, 1]$, and $f: [-1, 1) \rightarrow [0, 1] $ be $x \rightarrow x^2$
Then clearly the range is complete, but the domain is not complete, and $f$ is continuous, lastly we have that distance on $X$ is usually given by
$$|x - y|$$
Then the usual distance on $Y$ would be
$$|x^2 - y^2| = |x-y||x+y|$$
So in this case $d_y(f(x), f(x)) \le 2 d_X(x,y)$. We can consider a new metric then by considering $Y$ but with distance function
$$ d_Y(u,v) = \frac{1}{3} |u - v|$$
It follows then that $ d_X(x,y) \le d_Y(x^2, y^2) $
Does this look correct?
That is not correct, since $d_X(1,-1)=2$, whereas $d_Y\bigl(1^2,(-1)^2\bigr)=0$.
An example that works is $\tan\colon\left(-\frac\pi2,\frac\pi2\right)\longrightarrow\mathbb R$.