I have a particular question I am trying to solve.
True or false. Any surjective (onto) function $f:x \rightarrow x$ where $x$ is finite, must be bijective.
Since $x$ is finite, and since the domain and codomain are both the same, I am tempted to argue true, because I don't see how else each element in x could be hit. Am I wrong?
You are correct, as the comments point out. Two quick comments:
(1) You don't need induction, although you can use it if you want to. Given a map $f: X\rightarrow X$ which is surjective but not injective, you can build an injection of $\mathbb{N}$ into $X$ (fun exercise), so $X$ couldn't have been finite.
(2) Very weirdly, the converse need not hold (unless we assume the axiom of choice): it is consistent with ZF, the axioms of set theory minus choice, that there are infinite sets $X$ such that any onto map $X\rightarrow X$ must be a bijection. Such sets are called Dedekind-finite.
Yes because given $f:X \rightarrow X$ with $X$ finite:
Let $|x| = n \in \mathbb{N} $. For every $a\in x$, define $g(a)=|f^{-1}(\left \{ a\right \})|$, i.e. $g(a)$ is pre-image of the singleton set $\left \{ a\right \}$ under $f$. By surjectivity of $f$, we know that $\forall y\in x \ \ g(y)\neq 0$. Now define: $A=\left \{g(x):x\in x \right \}$. As $x$ is finite $A$ has the minimal element, say m. Now we know that (under assumption that $f$ is well-defined) that every element of $x$ is mapped to exactly one element of x, so in particular: $ n*m\Leftarrow n \Rightarrow m=1$, as $m\neq 0$. Similar argument for $\mathbb{E}(A)$ (Average value of elements of $A$) shows that average value equals to the value of minimal element, so all elements of $A$ are equal to $1$, so $f$ it's both injective and surjective, and the result follows.
