Mapping reductions for dummies I am having trouble understanding a mapping reduction and I would appreciate your help. Define

$\quad \begin{align}
  A_{TM} &amp;= \{ \langle M, w \rangle \mid M \text{ Turing machine}, w \in \mathcal{L}(M)\} \\
  S_{TM} &amp;= \{ \langle M,w \rangle \mid M \text{ Turing machine}, w \in \mathcal{L}(M) \implies w^R \in \mathcal{L}(M)\} \\
\end{align}$

and consider the reduction of $A_{TM}$ to $S_{TM}$ as follows.

Given $\langle M, w \rangle$ the following Turing machine $M'$ is defined:

M' on input x:  
  if x = 01 then accept  
  else run M on w and accept x if M accepts w


I don't understand the reduction entirely, this reduction is supposed to solve $S_{TM}$ using $A_{TM}$.  Why do I need to check if x = 01? There is no need to check anything about the reverse of $w$?  How is that covered by the reduction?
The idea of the reduction you give in your question is the following:

I) If $\langle M,w\rangle \in A_{tm}$, then $L(M')=\Sigma^*$.
II) If $\langle M,w\rangle \notin A_{tm}$, then $L(M')=\{10\}$.

thus, for any $w$, $\langle M,w\rangle \in A_{tm}$ iff $\langle M',w\rangle \in S_{tm}$.
In case (I), for any $w\in L(M')$ also $w^R\in L(M')$ hence $\langle M',w \rangle S_{TM}$ for any $w$. For case (II), there is no $w\in L(M')$ such that $w^R\in L(M')$ as well, thus for any $w$, $\langle M',w \rangle S_{TM}$.

Now for your question about accepting "10" no matter what: Assume we avoid this step. Then, in case (II), $L(M')=\emptyset$, hence $\langle M',w\rangle \in S_{tm}$ for any $w$, since the condition on $w,w^R$ vacuously holds. Now the mapping cannot map inputs  $\langle M,w\rangle \notin A_{tm}$ to inputs outside $S_{TM}$ (since none exists).

The value "10" is arbitrary. Any other non-palindromic string will do.