Difference in 1-1 vs many-1 Reducibility Let B be many-one reducible to A i.e. $B \leq_m A$ if there is a computable function $f$ s.t. for all $x\in \mathbb{N}$:
$$
x \in B \iff f(x) \in A
$$
If the function $f$ is one-one (injective) then B is said to be one-one reducible to A  i.e. $B \leq_1 A$. 
As far as most of the papers I have read, these two measures of reducibility coincide completely. However I keep reading mentions of the fact that one-one and many-one reducibility are actually two different metrics.
Can anyone point me to an example of c.e sets A and B s.t. $B \leq_m A$ but $ B \not \leq_1 A$? In particular with $B$ not being computable?

Many-one and one-one reducibility are very different.
Here's one way to see this: for a function $g:\omega\rightarrow\omega$ and a sequence $S\in2^\omega$, let the $g$-spread of $S$ be the sequence $Spread_g(S)$ gotten by "stretching" the $n$th bit of $S$ by a factor of $g(n)$. So e.g. if $g: x\mapsto x$ and $S=(0,1,0,1,0,1,...)$, then $Spread_g(S)$ is $$(0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0,0,0,0,0,1,1,1,1,1,1,...).$$
Note that we always have $Spread_g(S)\le_m S$ whenever $g$ is computable - just map each number in the $n$th "$g$-block" to $n$. However, this is very noninjective. The point is in general that there is no way to fix this - if $g$ is computable and infinitely often greater than $1$, then we can find an $S$ such that $Spread_g(S)\not\le_1 S$.
The intuition behind this is the following. Take $g: x\mapsto 2$, and $S$ some complicated set. If $f$ is to be a $1$-reduction of $Spread_g(S)$ to $S$, what should $f(3)$ be? We know that $3\in Spread_g(S)\iff 2\in Spread_g(S)$, so $f$ needs to send $3$ to something in the $k$th $g$-block (that is, an element of $\{2k-1, 2k\}$) such that $S(k)=S(2)$ (remember, $3$ is in the second $g$-block - hence the $2$). 
But the next such block could be very far away! In particular, let $S$ be any sequence whose principle function (= take $n$ to place in $S$ where we see $n$th $1$) is not computably dominated; then no computable $f$ can always "look far enough" to find an un-mapped to $g$-block, so either isn't even an $m$-reduction or isn't injective.
Basically, $1$-reductions are very sensitive to "padding out" the sets involved, whereas many-one reductions aren't.
