I am working on old qualifier problems as a review, and I came across this one:
Suppose there exists a continuous surjection $f:X_1 \mapsto X_2$, where $(X_1,d_1),(X_2,d_2)$ are metric spaces, such that $d_1(x,y) \leq d_2(f(x),f(y))$ for all $x,y \in X_1$. Prove or give a counterexample:
(a) If $X_1$ is complete, then so is $X_2$.
(b) If $X_2$ is complete, then so is $X_1$.
I believe I have solved (a), but I would appreciate your answer so I can check. For (b), would it work to use $X_1=(0,1]$, $X_2=[1,\infty)$, and $f(x)=1/x$ as a counterexample? Do you know any other examples?
Your example for (b) is fine and reminds one that completeness is really a poroperty of metric spaces, not a topological property (as $x\to \frac1x$ is a homeomorphism). The idea behind your example is of course to take a space with a nonconverging Cauchy sequence and to transport its nonexistent limit out of reach for the target space. Insofar, your example is the mother of all examples, I suppose.
For (a), let $\{y_n\}_n$ be a Cauchy sequence in $X_2$. Then by surjectivity there exist $x_n\in X_1$ with $f(x_n)=y_n$. By the given condition, $\{x_n\}_n$ is a Cauchy sequence. As $X_1$ is complete, $x_n\to x$ for some $x\in X_1$. Then $y_n\to f(x)$ by continuity of $f$. (According to the emphasis, we made good use of all the given facts).