How do I show continuous image of complete space is complete?

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If $f:X \rightarrow Y$ is uniformly continuous and $X$ is complete then so is $f(X)$?

My idea is that this is not true. $f :\mathbb{R} \rightarrow (-\pi/2,\pi/2)$, by $f(x)=tan^{-1}(x)$.

I think this result is true only if $f(X)=Y$.

Can anyone suggest some comments on this idea?