Surjectivity of sign function $\epsilon:S_n\rightarrow\{-1,1\}$ where $S_n$ is the symmetric group on $n$ symbols

Question

It is a proposition in Dummit and Foote book that $\epsilon$ as mentioned in the title is a surjective homomorphism. It is based on the reasoning that the sign of the identity permutation is 1 and sign of any transposition is -1. But in case of $S_1$ we don't have a transposition. So how do we ensure surjectivity of $\epsilon$ in this case. I mean which is the permutation in $S_1$ which maps to -1.

Answer 1

Since $\lvert S_1\rvert=1$ and $\bigl\lvert\{1,-1\}\bigr\rvert=2$, there is no surjective map from $S_1$ onto $\{1,-1\}$. That statement only holds when $n>1$.