Number of different bijections between a set $S$ with $n$ elements and itself

Question

I am trying to solve exercise 2.1 in Aluffi's textbook "Chapter 0":

How many different bijections are there between a set $S$ with $n$ elements and itself?

Here is my attempt. My "non-rigorous" answer is that I can make the function injective by choosing $n$ elements for where to make the first element of $S$, $n-1$ possibilities for the second, and so forth, giving $n!$ functions. Then this is a function injective function from $S$ to $S$ which is finite, so it is immediately surjective. (I don't think surjectivity just "follows" without invoking this kind of obvious, though it sounds obvious that it's surjective, though it's somewhat tough to prove.) Here is my attempt at making the argument rigorous, but I'd appreciate feedback on this above sketch as well.

Since there are $|n|$ choices for where to send each $s_i \in S$, there are exactly $n^n$ functions $f: S \to S$. If $f(s_1) = f(s_2)$ for $s_1 \neq s_2$, then $\mathrm{Im}(f) \not \subset S$, i.e., its cardinality is $< n$, so it is not surjective; furthermore, it is not injective. There are $n!$ mappings $S \to S$ with no repeated inputs or repeated outputs, which are the only bijections $S \to S$.

How does this look? Are there ways to make it more rigorous?

Answer 1

Alternative (much less elegant) approach.

Label the elements in $S:~~~s_1, s_2, \cdots, s_n$.

Let $P$ denote the $(n^n)$ possible mappings from $S$ onto itself.
Let $Q$ denote the subset of $P$ that consist of all bijections from $S$ onto itself.

Let $N = \{1,2,3,\cdots, n\}.$
Let $A$ denote the $(n^n)$ possible mappings from $N$ onto itself.
Let $B$ denote the subset of $A$ that represents all of the bijections from $N$ to $N$.
Let $C$ denote the subset of $A$ that represents all of the ways of permuting the numbers in $N$.

Per this Wikipedia article:

• $C$ has exactly $(n!)$ elements.
• $C = B \implies B$ has exactly $(n!)$ elements.

Consider the mapping $\lambda: P \to A$ defined as follows:
For any $p \in P, ~a = \lambda(p)$ is defined as follows:
$a(i) = j \iff p(s_i) = s_j.$

Take any $q \in Q$, and let $a = \lambda(q).$ Since $q$ is a bijection, you know that

• For each element $s_j \in N$, there is exactly one element $s_i \in N$ such that $q(s_i) = s_j.$

Per the definition of $\lambda$, this implies that

• For each element $j \in N$, there is exactly one element $i \in N$ such that $a(i) = j.$

Therefore, $a$ is a bijection from $N$ to $N$.
Therefore, $a \in B$.
Therefore, $\lambda(Q)~$ [i.e. the set of all $\lambda(q)$, where $q \in Q$] is a subset of $B.$

Let $~~b~~$ be any element in $B$, which implies that $b$ is a bijection from $N$ onto itself.

Construct the element $f$ in $P$ as follows:
$f(s_i) = s_j \iff b(i) = b(j).$

Because $b$ is a bijection, you know that for each element $j$ in $N$ there is exactly one element $i$ in $N$ such that $b(i) = j.$

Because of the construction done to create $f$, you know that:

• For each element $s_j$ in $S$, there is exactly one element $s_i$ in $S$ such that $f(s_i) = s_j$. This implies that $f$ is a bijection from $S$ to $S$. Therefore, $f$ is in $Q$.

• $\lambda(f) = b$.

Therefore, $\lambda(Q) = B.$

Let $\lambda'$ represent the mapping $\lambda$ with its domain restricted to $Q$.

Then, since $\lambda(Q) = B, ~\lambda'$ is a surjection from $Q$ onto $B$.

Further, per the definition of $\lambda$, it is immediate that $\lambda$ is injective. That is, you can't have two distinct elements $f_1, f_2 \in P$ such that $\lambda(f_1) = \lambda(f_2).$

Consequently, $\lambda'$ inherits the property of being an injection. Therefore, $\lambda'$ is a bijection from the finite set $Q$ to the finite set $B$. Therefore, since $B$ has $(n!)$ elements, so does $Q$.

Answer 2

There are $n $ choices for the first element, $n-1$ for the second etc... So $n!$.

Personally I think it is obvious that a surjection, or injection, which is an endomorphism on a finite set, is a bijection.

(But some may question my rigor here, I guess.)