Prove a map X → X must be bijective

Question

Let X be a finite set. Does there exist a map $α : X → X$ such that $α$ is surjective, but not injective? $$ \text{No.} $$

$$\begin{align} \text{Surjective: }& \ \ {\displaystyle \forall y\in Y,\exists b\in X, y=\alpha(b)} \\\ \\ \text{Injective: }& \ \ {\displaystyle \forall x,y\in X,\alpha(x)=\alpha(y)\Rightarrow x=y} \\\ \\ \text{not Injective: }& \ \ {\displaystyle \forall y \in X, \exists x\in X,\alpha(x)=\alpha(y), x \ne y} \\\ \\ \end{align}$$ $$ \text{Noting that the map is to the same set } (Y=X), \\ \text{ we use the definition of surjective and not injective for proof by contradiction: } \\\ \\ \forall y\in X,\exists b,x\in X, y=\alpha(b),\alpha(x)=\alpha(y), x \ne y \\ \alpha(x)=\alpha(\alpha(b)) \\ $$

When I saw this problem it was immediately clear that the answer was no. By symmetry, a map to the same set must be injective if it is surjective. However, I am failing to see that symmetry in the definition of surjective and injective and exploit it to prove the fact.

Perhaps I must also include the definition of a function?

Could anyone explain how I can prove this, or at least give me a definition of injective such that, by switching $X$ and $Y$ it becomes the definition of surjective, or explain to me why that is not possible.

Answer 1

Your formulation of "not injective" is wrong and the formulation of "injective", $$\forall x,y\in X,\alpha(x)=\alpha(y)\Rightarrow x=y,$$ is sloppy.

The correct formulation of "injective" is

$$\forall x \in X: \;\forall y\in X:\;\alpha(x)=\alpha(y)\Rightarrow x=y$$

The "$:$" is only to improve the readibility of the formula.

The negation:

$$\lnot (\forall x \in X\!: \;\forall y\in X\!:\;\alpha(x)=\alpha(y)\Rightarrow x=y)\\ \equiv \exists x \in X\!: \;\lnot(\forall y\in X\!:\;\alpha(x)=\alpha(y)\Rightarrow x=y)\\ \equiv\exists x \in X\!: \;\exists y\in X\!:\;\lnot(\alpha(x)=\alpha(y)\Rightarrow x=y)\\ \equiv\exists x \in X\!: \;\exists y\in X\!:\;(\alpha(x)=\alpha(y)) \land (x\ne y)$$

So "not injective" is $$\exists x \in X, \exists y\in X,\alpha(x)=\alpha(y) , x\ne y$$ in your notation.

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I can't see a symmetry in the definition of injectivity and surjectivity. This "symmetry" of injectivity and surjectivity shoul be better called "equivalency". Only for finite sets surjectivity and injectivity are equvivalent. On infinite set they are no related. So I don't think that you will see a symmetry in the definitions.

Answer 2

We need an operational definition of "finite set". A (nonempty) set $X$ is finite if there is an $n\in{\mathbb N}$ and a bijection $\psi:\>X\to [n]$. It is therefore sufficient to prove the following:

Claim. If $f:\>[n]\to[n]$ is surjective then $f$ is injective.

Proof. The claim is true for $n=1$. Assume that it is true for $n$, and let a surjective $f:\>[n+1]\to[n+1]$ be given. We distinguish two cases:

(i) If $f(n+1)=n+1$ define $$g(x):=\left\{\eqalign{f(x)\quad&\bigl(f(x)\ne n+1\bigr)\cr 1\quad&\bigl(f(x)=n+1)\cr}\right.\qquad(1\leq x\leq n)\ .\tag{1}$$ Then $g:\>[n]\to[n]$ is surjective, hence injective by the induction hypothesis. It follows that the second line in $(1)$ is not called at all, so $f$ is injective as well.

(ii) If $f(n+1)=c\leq n$ define $$g(x):=\left\{\eqalign{f(x)\quad&\bigl(f(x)\ne n+1\bigr)\cr c\quad&\bigl(f(x)=n+1)\cr}\right.\qquad(1\leq x\leq n)\ .$$ It is easily checked that $g:\>[n]\to[n]$ is surjective, hence injective by the induction hypothesis. It then follows that $f$ is injective as well; in particular the values $c$ and $n+1$ are taken exactly once.