$A_{TM}= \{ \langle M,w\rangle \mid M$ is a TM that accepts $w\}$
$E_{TM}= \{ \langle M\rangle \mid L(M) = \emptyset \}$
The standard proof for the undecidability of $E_{TM}$ is given in this question, right?. Doesn't this automatically create a mapping reduction $f$ from $A_{TM}$ to $E_{TM}$ ? Whenever $w \in A_{TM}, f(w) \in E_{TM} $ obviously. The other way: $ f(w) \in E_{TM} \Rightarrow w \in A_{TM} $ seems true, although I suspect that's where the fundamental restriction lies. Can someone clarify this?
The proof given for $A_{TM}$ not being Turing reducible to $E_{TM}$ is the following:
Suppose for a contradiction that $A_{TM}$ $≤_{m}$ $E_{TM}$ via reduction $f$. This means that $w ∈ A_{TM} \iff f(w) ∈ E_{TM} $, which is equivalent to saying $w \notin A_{TM} \iff f(w) \notin E_{TM} $. Therefore, using the same reduction function $f$, we have that $\overline{A_{TM}} \le_m \overline{E_{TM}} $. However, $\overline{E_{TM}}$ is Turing-recognizable and $\overline{A_{TM}}$ is not Turing-recognizable.
I understand the proof above I think. It seems I am misunderstanding something fundamental about mapping reductions, or that I am confused about some step in the reduction from $A_{TM}$ to $E_{TM}$ in the standard proof.