$f$ continuous and surjective, $d_1(a,b)\le d_2(f(a),f(b))$, $X$ complete implies $Y$ complete Let $(X,d_1)$ and $(Y,d_2)$ be metric spaces. Let $f : X \to Y$ be continuous and surjective. Suppose $d_1(a,b)\le d_2(f(a),f(b))$ for all $a,b\in X$. How can we show that if $X$ is complete then $Y$ is complete?
 A: Hint: 


*

*Consider any Cauchy sequence $y_n\in Y$

*Let $x_n$ be a sequence  s.t. $f(x_n) = y_n$ for all $n$. Why for any $y_n$ such $x_n$ exist?

*Prove that $x_n$ is Cauchy, and let $x$ be its limit

*show that $y_n \to f(x)$
A: Let $(y_n)=f(x_n)$ be Cauchy in $Y$. Then for any $\epsilon >0$, there is an $M>0$ such that for all $m,n>M$, we have $d_1(x_m,x_n)\leq d_2(y_m,y_n)<\epsilon$. This means $(x_n)$ is Cauchy in $X$ and converges to some $x\in X$ by completeness of $X$. Hence $(y_n)$ converges to $f(x)$ by continuity of $f$.
A: Proof:
Let $\{y_n\}_{n\in\mathbb{N}}$ be a Cauchy sequence,
$$\forall \epsilon>0.\;\exists N\in\mathbb{N}.\;\forall n,m\ge N.\;d_2(y_n,y_m)<\epsilon$$  Since $f$ is surjective, we can find $\{x_n\}_{n\in\mathbb{N}}$ such that $\forall n\in\mathbb{N}.\;f(x_n) = y_n$.
$$\Longrightarrow\forall m,n\ge N.\;d_1(x_m-x_n)\le d_2(f(x_m)-f(x_n)) = d_2(y_m-y_n)<\epsilon.$$
$\{x_n\}_{x\in\mathbb{N}}$ is Cauchy, it converges in $X$ becasue $X$ is complete. 
Denote $\xi: = \lim_{n\to\infty} x_n$  
Since $f$ is continuous, it is sequentially continuous, that is
$$\lim_{n\to\infty} y_n=\lim_{n\to\infty} f(x_n) = f(\lim_{n\to\infty}x_n) = f(\xi)  $$
Therefore $\{y_n\}_{n\in\mathbb{N}}$ converges. Since $\{y_n\}_{n\in\mathbb{N}}$ is chosen arbitrary, every Cauchy sequence in $Y$ converges $\Longrightarrow Y$ is complete.
