$S$ is a finite set and $f:S\rightarrow S$ is surjective. Show $f$ is injective.
Proof) Suppose $f$ is not injective. Then $\exists s_1\neq s_2$ with $f(s_1)=f(s_2)$. Then since $S$ is finite, $|f(S)|\leq |S|-1$.($|\cdot|$ denotes the number of elements of a set.) Also, since $f$ is surjective, $f(S)=S$, so $|f(S)|=|S|$. Then $|S|\leq|S|-1$, a contradiction.
I was wondering if this is a correct argument.
The proof is correct, although perhaps you could justify the claim that $|f(S)|\leq |S|-1$ more clearly. In particular, since $S$ is finite suppose that $f(a_i)=b_i$ for each $1\leq i\leq |S|$, where the $b_i$ are a permutation of the $a_i$. If $f$ is not injective, without loss of generality let $f(a_1)=f(a_2)$, whence $b_1=b_2$, so the set $$f(S)=\{b_1,b_2,\dots,b_{|S|}\}$$ contains at most $|S|-1$ distinct elements.
