Let $(X,d_1)$ and $(Y,d_2)$ be two metric spaces, $f:X \to Y$ be continuous surjection and isometric. Prove that if $(X,d_1)$ is complete, than $(Y,d_2)$ is also complete m.s..
My work: Let $(X,d_1)$ be a complete metric space, which means that every Cauchy sequence in $X$ has a limit that is also in $X$. And let $f$ be an isometry, which is a transformation that maps elements to the same or another metric space, such that the distance between the image element in the new metric is equal to the distance between the elements in the original metric space.
From the definition for surjection, we know that, for every $b$ in $Y$, there exists an $a$ in $X$ such that $f(a)=b$. And I don't know where to from here.
To show that $(Y,d_2)$ is complete, we let $(y_n)_{n=1}^\infty$ be a Cauchy sequence in $Y$ and we show that it has a limit in $Y$. As $f$ is surjective, for every $y_n$ there is some $x_n$ such that $f(x_n) = y_n$. Now you can show that $(x_n)_{n=1}^\infty$ is Cauchy. As $(X,d_1)$ is complete, there is some $x\in X$ such that $(x_n)_{n=1}^\infty$ converges to $x$. Now, define $y=f(x)$ and show then show that $y$ is the limit of $(y_n)_{n=1}^\infty$.
To show that $(x_n)_{n=1}^\infty$ is Cauchy and that $(y_n)_{n=1}^\infty$ converges to $y$ you can use that $$ d_1(x_n,x_m) = d_2(f(x_n),f(x_m)), $$ for all $n,m$.