 Solution

Because $f$ is a bijection and $e$ is a metric,is nonnegative, symmetric, obeys the triangle inequality, and vanishes exactly when $x=y$. Hence it is a metric. Its open sets are precisely the inverse images under $f$ of $e$-open sets in $N$. Since $f:(M,d)\to (N,e)$ is a homeomorphism, these are exactly the $d$-open sets. Thus $d^{\prime }$ and $d$ are equivalent metrics.

Leta homeomorphism $\mathbb{R}\to (0,1)$, and defineThis metric induces the standard topology on $\mathbb{R}$. However, the sequence $x_n=n$ is $d^{\prime }$-Cauchy because $h(n)\to 1$. If it converged to $x\in \mathbb{R}$, then continuity of $h$ would give $h(x)=1$, impossible. Therefore $(\mathbb{R},d^{\prime })$ is not complete.