Two metrics induce the same topology, but one is complete and the other isn't

The metric space $\{\frac {1}{n} \mid n\in \mathbb{N} \}$ with the usual metric is incomplete (since we don't have zero), and it has the discrete topology.

the same space with the metric $d(x,y)=1 \iff x\neq y$ also has a discrete topology but is complete, since any cauchy sequence will eventually be constant.

The interval $M:=]-1,1[$ with the usual line element $ds:=|dx|$ is an incomplete metric space: The sequence $x_n:=1-{1\over n} \ (n\to\infty)$ converges in $\mathbb R$, so it is a Cauchy sequence, but it diverges in $M$. On the other hand, the "hyperbolic metric" defined by $ds:=|dx|/(1-x^2)$ induces the same topology on $M$ but is complete. The latter statement needs of course a proof. Suffice it here to say that now the endpoints $\pm1$ are "infinitely far away".

Look for a homeomorphism $f: \mathbb R \to \mathbb R_+$ (you should know a good one). Then take the usual metric on $\mathbb R$ and use it to define a metric $d$ on $\mathbb R_+$ by $d(x,y) = |f(x)-f(y)|$ (*). Now what can you say about the relationship between $d$ and the usual metric on $\mathbb R_+$?

EDIT: Made the question here correct and more illustrative.

So the identity $\iota: (\mathbb R_+,d) \to (\mathbb R_+,|\cdot|)$ is a homeomorphism. Now just ask yourself which property, that continuous functions between metric spaces can have, $\iota$ doesn't have? Can you formulate a theorem about what property a homeomorphism must have to preserve completeness?

(*): Pullback and pushforward might be of interest.