Suppose $X$ is a finite set and $f : X \to X$ is a function. Then $f$ is injective if and only if $f$ is surjective.

Solution 1:

Your proof is a little informal. Normally this wouldn't be a problem, but the issue is that the formal version of your "counting" argument is precisely what you wish to prove. The proof is not as simple as one would initially be lead to believe.

Let's prove the following statement. Your desired statement will follow.

Let $n\in\mathbb{N}$ and let $f:\{1,\ldots,n\}\to\{1,\ldots,n\}$ be a function. Then $f$ is injective iff $f$ is surjective.

Proof:

We use induction. This clearly holds whenever $n=1$. Fix $n\in\mathbb{N}$ and suppose that every function from $\{1,\ldots,n\}$ into $\{1,\ldots,n\}$ is injective iff it is surjective. Let $f:\{1,\ldots,n+1\}\to\{1,\ldots,n+1\}$ be a function.

Suppose $f$ is injective. Denote $m:=f(n+1)$, and observe that since $f$ is injective, the function $g:\{1,\ldots,n\}\to\{1,\ldots,n+1\}\setminus\{m\}$ given by $g(k)=f(k)$ is well-defined and one-to-one. Define the injection $h:\{1,\ldots,n+1\}\setminus\{m\}\to\{1,\ldots,n\}$ by $$ h(k) = \begin{cases} k & \text{if}\ k<m, \\ k-1 & \text{if}\ k>m. \end{cases} $$ Due to the fact that the composition of injective functions is injective, we have that $h\circ g:\{1,\ldots,n\}\to\{1,\ldots,n\}$ is one-to-one. By the induction hypothesis, $h\circ g$ is onto and hence $h$ is onto. Therefore $g=h^{-1}\circ(h\circ g)$ is a composition of surjective functions, which means that it is surjective. Consequently $f$ is surjective.

(I'll leave the converse to you). $\square$

As far as the examples go, the exponential function works perfectly. However, a logarithmic function doesn't work because it isn't defined on all of $\mathbb{R}$ (and it is injective). To fix this, you could define $f:\mathbb{R}\to\mathbb{R}$ by $f(x)=\ln|x|$ if $x\ne0$ and $f(0)=0$.

Edit: Actually, one can prove that this statement never holds whenever $X$ is infinite. Suppose $X$ is an infinite set. Then there is a countably infinite subset $\{d_1,d_2,\ldots\}$ of $X$. (Caution: This uses the axiom of choice.) Define $f_i,f_s:X\to X$ by $$ f_i(x)=\begin{cases} d_{i+1} & \text{if}\ x=d_i\ \text{for some}\ i, \\ x & \text{otherwise}, \end{cases} \quad\text{and}\quad f_s(x)=\begin{cases} d_{i-1} & \text{if}\ x=d_i\ \text{for some}\ i>1, \\ x & \text{otherwise}. \end{cases} $$ Then $f_i$ is an injection and $f_s$ is a surjection, but neither is a bijection.

Solution 2:

The idea of the proof you gave is correct, although it is written slightly informal.

An exponential function would indeed work: define $f: \mathbb{R} \to \mathbb{R}: x \mapsto e^x$. This function is clearly injective (it is strictly increasing), but $e^x > 0$ where $x \in \mathbb{R}$, so the mapping is not surjective.

For the other example you have to provide, define $g: \mathbb{R} \to \mathbb{R}: x \mapsto x^3 - x$. Clearly, this function is not injective ($f(0) = f(1)$), but it is surjective