Mapping Reductions to Complement of A$_{TM}$
Let's take $L_{\emptyset}=\{ \langle M \rangle \mid L(M) = \emptyset \}$, that is, all the machines that accept no word (i.e., whose language is empty).

Now we show the reduction $\overline{A_{TM}} \le L_\emptyset$. The reduction is done by taking an input $(\langle M \rangle,w)$ of $\overline{A_{TM}}$ and converting it into an input ${\langle \tilde M \rangle}$ for $L_\emptyset$ such that

$$(\langle M \rangle,w) \in \overline{A_{TM}} \quad\text{  iff  }\quad \langle \tilde M \rangle \in L_{\emptyset}$$

Given $(\langle M \rangle,w)$ we can construct $\tilde M$ in th following way. $\tilde M$ on input y does the following:


deletes the tape
writes $w$ on the tape
runs $M$ on $w$, and performs the same (if $M$ accepts, $\tilde M$ accepts as well).


Convince yourself it is possible to construct the coding of $\tilde M$ from the coding of $M$ and from $w$. Now let's verify that this reduction is valid:


If $(\langle M \rangle,w) \in \overline{A_{TM}}$ then $M$ either rejects or doesn't halt. If so, then also $\tilde M$ doesn't accept $y$, for any input $y$. This means $L(\tilde M) = \emptyset$ thus $\langle \tilde M \rangle \in L_{\emptyset}$.
If $(\langle M \rangle,w) \notin \overline{A_{TM}}$ then $M$ accepts $w$, thus $\tilde M$ accepts $y$ (for every $y$). It follows that $L(\tilde M)=\Sigma^*$ which implies that $\langle \tilde M \rangle \notin L_{\emptyset}$.


The "iff" condition holds and we successfully mapped an input of $\overline{A_{TM}}$ into an input of $L_\emptyset$. In this case we say we reduced $\overline{A_{TM}}$ to $L_\emptyset$. That is, if we can solve $L_\emptyset$, we can also solve $\overline{A_{TM}}$ by first converting the input and then running the algorithm that solves $L_\emptyset$ on the converted input.