How prove the function could extend continuously to completion of metric spaces is equivalent to the function maps Cauchy sequence to Cauchy sequence?

Question

The question comes from my friend. We had a long discussion but only got a complex answer. I wish to get a simply answer.

To be more precise,

Given two metrix spaces $ X $ and $ Y $, a (continuous) function $f \colon X \to Y$. The following conditions is equivalent.

1. $ f $ maps Cauchy sequence to Cauchy sequence;
2. for the completion $\hat{X}$ and $\hat{Y}$, there exists an continuous extension $\hat{f} \colon \hat{X} \to \hat{Y}$, that means, the restriction of $\hat{f}$ on $X$ is equal to $f$, and $\hat{f}$ is continuous.

My main doubts about this issue is it might not maintain Cauchy continuity. Because imaging the sequence on sequence is difficult for me.