Suppose we have a partial computable injective function $f: \mathbb N\to \mathbb N$ whose image is computable. I'm trying to show that then the image of every computable set is computable, and that this is false for non-injective functions (by a counterexample).
Suppose $X$ is computable and one wants to check if $y$ lies in $f(X)$. The problem asks to construct an algorithm of how to do this. What I can think of is this: the algorithm accepts $y$. Since $im(f)$ is computable, we one can tell whether $y\in im(f)$. If this is false, then $y\notin f(X)$ either. Suppose $y\in im(f)$. Then for some $x\in dom(f)$, $f(x)=y$. Now it would be natural to find such an $x$ (which would be unique due to injectivity). (And then one would be able to tell whether $x\in X$ or not.) But how to find an algorithm/program doing this? The domain of $f$ may be infinite, so it may be not possible (in finite time) to check for every $x\in dom(f)$ if $f(x)=y$. Also I don't see how the computability of $f$ can be used here.
As for the counterexample, once I understand how to find $x$ with $f(x)=y$, the most trivial counterexample should work, like $\{1,2\}\to \mathbb N, 1,2\mapsto 1$. Is that right?