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Suppose that $d_{1}$ and $d_{2}$, are equivalents metrics onto a non-empty set $X$, and consider a function $f:X\to Y$, where $(Y,d)$ is a arbitrary metric space. Show that $f$ is continuous onto $(X,d_{1})$ iff $f$ is continuous in $(X,d_{2})$.

My approach: We know that, two metrics $d_{1}$ and $d_{2}$ in a space $X$ are equivalents when, the identity $i_{1,2}:(X,d_{1})\to (X,d_{2})$ is a homeomorphism with $i_{1,2}(x)=x$, for $x\in X$. Consider $f:X\to Y$, where $(Y,d)$ is a metric space. (My idea, is use the function $i_{1,2}$ of any way) If $f:(X,d_{1})\to(Y,d)$ is continuous, then $f'=f\circ i_{2,1}:(X,d_{2})\to (Y,d)$ is continuos, since $f$ and $i_{1,2}$ are continuous. We just considering $f'\equiv f:(X,d_{2})\to (Y,d)$. Analogue in the other direction

Ahna Akbar
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First show (if this isn't already your definition) that two metrics $d_1$ and $d_2$ on $X$ are equivalent if and only if the sets in $X$ which are open with respect to $d_1$ are also open with respect to $d_2$, and vice versa.

Then show (if this isn't already the definition you're using) that a function between metric spaces $f: X \rightarrow Y$ is continuous if and only if whenever $V \subseteq Y$ is open in $Y$, the preimage $f^{-1}V$ is open in $X$.

D_S
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