Proof that the sign is a homomorphism

Proof

We concatenate decompositions into transpositions. Products in the symmetric group $S_n$ are composites of maps, and $n \geq 2$.

1. Let $\sigma, \tau \in S_n$. By the decomposition into transpositions, write $\sigma = \tau_1 \cdots \tau_m$ and $\tau = \tau'_1 \cdots \tau'_l$, where each $\tau_i$ and each $\tau'_j$ is a transposition.
2. By the definition of the sign, $\operatorname{sgn}(\sigma) = (-1)^m$ and $\operatorname{sgn}(\tau) = (-1)^l$. This does not depend on the chosen decompositions, because the sign is well defined.
3. By associativity, $\sigma \tau = \tau_1 \cdots \tau_m \tau'_1 \cdots \tau'_l$. This is a product of $m + l$ transpositions.
4. By the definition of the sign again, $\operatorname{sgn}(\sigma \tau) = (-1)^{m+l} = (-1)^m (-1)^l = \operatorname{sgn}(\sigma) \operatorname{sgn}(\tau)$.
5. The set $\{1, -1\}$ is a group under multiplication, and step 4 says that $\operatorname{sgn}$ preserves products. So $\operatorname{sgn}$ is a group homomorphism.
6. For surjectivity, the transposition $(1\ 2)$ is a product of one transposition, so $\operatorname{sgn}((1\ 2)) = -1$. This uses $n \geq 2$.
7. The identity $e$ equals $(1\ 2)(1\ 2)$, a product of two transpositions. So $\operatorname{sgn}(e) = (-1)^2 = 1$.
8. By steps 6 and 7, both $1$ and $-1$ are values of $\operatorname{sgn}$. So $\operatorname{sgn} \colon S_n \to \{1, -1\}$ is a surjective group homomorphism.