functional analysis - On injective, bijective, surjective

Consider a function $f: \mathcal{A}\rightarrow \mathbb{R}$ where $\mathcal{A}\equiv \{a_1,a_2,a_3,a_4\}\subset \mathbb{R}$ .

We know that $f$ is injective, i.e., $f(a_j)\neq f(a_k)$ $\forall a_j\neq a_k$.

I would like your help with the following: given that the domain of $f$ is finite (and hence the image set of $f$ is finite), does this mean that $f$ being injective implies $f$ being bijective? If not, can you give an example of $f$ with domain $\mathcal{A}$ that is injective but not bijective?

Answer

Quite the opposite, in fact: there are no bijective (or even surjective) functions from a finite set to an infinite set (this a perfectly reasonable definition of "infinite set"). For an explicit example, take the function $f: \mathcal{A}\to\mathbb{R}$ that sents $a_1$ to $1$, $a_2$ to $2$, $a_3$ to $3$, and $a_4$ to $4$. This is clearly injective, but clearly not bijective: nothing is sent to $5$. Any other injective function $\mathcal{A}\to\mathbb{R}$ would work equally.

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However, for the question that I suspect you meant to ask (at any rate, it's the closest thing to your statement that is true): if $f: A \to B$ is a function such that $A$ and $B$ are finite sets, and $|A| = |B|$, then the following are equivalent:

1. $f$ is injective.


2. $f$ is surjective.


3. $f$ is bijective.

To see this, note that clearly (3) implies both (1) and (2), and that if we have both (1) and (2), then we have (3), so we merely need to show that (1) implies (2), and (2) implies (1). For the former, note that if $f$ is injective, then there are $|A|$ distinct elements in its image. But $|A| = |B|$, so all elements of $B$ are in the image of $f$, so $f$ is surjective. For the latter, note that if $f$ is surjective, then all elements of $B$ are in its image, so there are $|B|$ distinct elements in its image, each of which must have at least one element of $A$ sent to it, but since $|B| = |A|$, each must have exactly one sent to it, and so no two elements of $A$ are sent to the same element of $B$, hence $f$ is injective.