A section/splitting of a homomorphism. Let $\epsilon : S_n \rightarrow \{\pm 1\}$ take each permutation to its sign. Then $\epsilon$ is an extension of $A_n$ by $Z_2$. Moreover, setting $s(-1)=\tau \in S_n$ determines a section of $\epsilon$ if and only if $\tau$ has order 2 and is an odd permutation. 
The definition of a section is,

A section, or splitting, for $f: G \rightarrow K$ is a homomorphism $s: K \rightarrow G$, such that $f \circ s$ is the identity map of K. 

I just wanted to be sure...when they say "setting $s(-1)=\tau \in S_n$ determines a section", they just mean that the homorphism that takes -1 to a permutation that has order 2 and is odd...and takes 1 to a permutation that is not of order 2 and is even. Is that correct? We can only say that "s(w)=v determines a section" if there are only two elements to consider, right?
Thanks in advance 
 A: You are essentially correct.  A few comments:


*

*A permutation that has order 2 must be odd.

*The splitting map is a homomorphism of groups, so it must carry identity to identity.  Hence, the splitting map is completely determined by where it sends the only non-identity element, $-1$.

*The image of the splitting, $s(\{\pm 1\}) \subseteq S_n$ is a little isomorphic copy of $\mathbb Z_2$ inside $S_n$.

A: The point is that to define a homomorphism $s : \{-1,1\} \to S_n$ you just need to chose $s(-1)= \tau$ because we necessarily have $s(1) = \textrm{Id}$ (because $s$ is a homomorphism).
You cannot just choose any $\tau \in S_n$ for $s$ to be a homomophism, you need to have $\tau^2 = s(-1)^2 = s((-1)^2) = s(1) = \textrm{Id}$. Conversely, you can check that for any $\tau \in S_n$ such that $\tau^2 = \textrm{Id}$, you can define a homomorphism $s : \{-1,1\} \to S_n$ (by setting $s(-1) = \tau$ and $s(1) = \textrm{Id}$).
Moreover, if $s$ is a section of $\epsilon$ we also need $-1 = \epsilon(s(-1)) = \epsilon(\tau)$. So if $\tau$ is an odd permutation of order $2$. Conversely, any such permutation $\tau$ defines a section $s$ (by setting $s(-1) = \tau$ and $s(1) = \textrm{Id}$).
A: Well, $s:C_2\to S_n$ must also be a homomorphism, so it cannot simply take $+1\in C_2$ to just any even permutation of $S_n$, $1$ must be taken to the identity. But given any odd permutation $\tau\in S_n$ of order $2$, there is indeed a homomorphism $s_\tau:C_2\to\langle\tau\rangle\le S_n$, and ${\rm sgn}\circ s_\tau={\rm id}_{C_2}$, so this is a section of the sign homomorphism $S_n\to C_2$.
In general, you can think of a section of $f:G\to G/N$ as a copy of $G/N$ existing within $G$. They are not always present in group extensions, only split extensions.
