Is the Sign of a Permutation Zero if and Only if It's Not a Bijection?

[autre]

Homework Statement

$sgn(\sigma)=0\iff\sigma$ is not a bijection.

The Attempt at a Solution

$(\rightarrow)$Let $sgn(\sigma)=0$. Then, $\Pi_{1\leq i<j\leq n}\frac{\sigma(j)-\sigma(i)}{j-i}=0.$. For some $i$ and $j$, $i\neq j$, $\sigma(i)=\sigma(j)$. Thus, $\sigma$ is not an injection.

$(\leftarrow)$ $\sigma$ not a bijection $\rightarrow$ it is not an injection, same argument as above, $sgn(\sigma)=0$.

I think I went wrong somewhere. Any ideas?

 

[mighty2000]

Your approach is generally correct, but there are a few minor mistakes. First, when you say "for some i and j, i\neq j , \sigma(i)=\sigma(j)", you should specify that i and j are distinct elements in the range of \sigma, not just any two integers. This is because the product only covers distinct pairs of elements in the range.

Secondly, when you say "thus, \sigma is not an injection", you should clarify that this is because \sigma is not one-to-one. This is important because a function can be non-injective without being a bijection (i.e. it could be onto), so it's important to specify what property of \sigma is not satisfied.

Other than that, your reasoning is correct and your proof is complete. Good job!