Is a set $B = \{y, \exists x \in A, f(x)=y\}$ recursive if A is a recursive set and f is a $N->N$ total computable function?
In general, no, we can't conclude that $B$ is recursive. We can choose $f$ so that $B$ is the halting problem, for instance.

To do that, choose any TM $M_0$ which halts on the empty input.

We define $f(x)$ as follows. Let $M,n$ be such that $\langle M, k \rangle = x$ where
$\langle -, - \rangle$ is any computable pairing function (bijection).
Then, run $M$ on empty input for $k$ steps. If that halts, return $\langle M \rangle$, otherwise return $\langle M_0 \rangle$.

It is then easy to verify that, using such $f$, we have

$$
\begin{array}{ll}
B &amp;= \{ \langle M' \rangle \ |\ \exists M,k. f(\langle M,k \rangle)= \langle M' \rangle \}
\\
&amp;= \{ \langle M' \rangle \ |\ \exists k. M' \mbox{ halts in $k$ steps } \}
\\
&amp;= \{ \langle M' \rangle \ |\ M' \mbox{ halts} \}
\end{array}
$$



With some more effort, we can even make $f$ injective, yet obtain the same $B$ (halting problem).

Indeed, a well-known result states that any RE nonempty set $A$ is the image of some total computable function $g: \mathbb N \to \mathbb N$. For the halting problem $A$ is infinite and RE, so there is such a $g$. Exploiting $g$ we can define $f(x)$ as $g(y)$ where $y$ is the least number such that $\{g(0),g(1),\ldots,g(y)\}$ contains exactly $x+1$ (non-duplicate) elements.

Note that $y$ exists since the image of $g$ ($A$) is infinite.

In such way the image of $f$ is the same as the one of $g$, i.e. $A$, but $f$ is now injective by construction. ($f$ is not surjective, of course)