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\begin{document}

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\section*{bijective proof}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{combinatorics}{}\paragraph*{{Combinatorics}}\label{combinatorics}

[[!include combinatorics - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{examples}{Examples}\dotfill \pageref*{examples} \linebreak
\noindent\hyperlink{binomial_identity}{Binomial identity}\dotfill \pageref*{binomial_identity} \linebreak
\noindent\hyperlink{binomial_identity_2}{$q$-binomial identity}\dotfill \pageref*{binomial_identity_2} \linebreak
\noindent\hyperlink{counting_trees_cayleys_theorem}{Counting trees: Cayley's theorem}\dotfill \pageref*{counting_trees_cayleys_theorem} \linebreak
\noindent\hyperlink{vandermonde_identity}{Vandermonde identity}\dotfill \pageref*{vandermonde_identity} \linebreak
\noindent\hyperlink{stirling_number_identities}{Stirling number identities}\dotfill \pageref*{stirling_number_identities} \linebreak
\noindent\hyperlink{polynomial_identities}{Polynomial identities}\dotfill \pageref*{polynomial_identities} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

In enumerative [[combinatorics]], a ``bijective proof'' refers to a basic method of counting the number of structures of a certain type supported on a finite set of underlying points, by analyzing structure in two different ways. It is really a special case of ``[[categorification]]'': an identity $a = b$ where $a$ and $b$ are [[natural number]] expressions is proved by taking underlying cardinalities of a [[bijection]] $A \cong B$ between sets of structures, hence the name. (If taking cardinalities is an example of ``decategorifying'', then going the other way from $a = b$ to $A \cong B$ would be ``categorifying'' or promoting an equation to an isomorphism.)

In some sense this is what the theory of [[species]] is all about: formalizing the operations adhering to bijective proofs. Often bijective proofs give an impression of elegance or enlightenment: rather than prove natural number or polynomial identities through acts of symbolic manipulation, one \emph{sees} the identity by drawing pictures of structures that the two sides of the identity are counting.

\hypertarget{examples}{}\subsection*{{Examples}}\label{examples}

For the moment we'll just jot down some possible examples for illustrating the method; these are to be filled in later, or to be linked to other nLab articles where the bijective proof is given.

\hypertarget{binomial_identity}{}\subsubsection*{{Binomial identity}}\label{binomial_identity}

See for now [[binomial theorem]].

\hypertarget{binomial_identity_2}{}\subsubsection*{{$q$-binomial identity}}\label{binomial_identity_2}

For more details, see the \href{https://en.wikipedia.org/wiki/Gaussian_binomial_coefficient}{Wikipedia article}.

Algebraically, one may introduce the $q$-binomial coefficients $\binom{n}{k}_q$ by introducing a $q$-affine plane (or quantum affine plane, to be connected with [[quantum affine algebra]]) and its associated (non-commutative) algebra

\begin{displaymath}
k[x, y]_q = k\langle x, y \rangle/(y x = q x y)
\end{displaymath}
where $q \in k$ is a parameter, and then writing $(x + y)^n = \sum_{k = 0}^n \binom{n}{k}_q x^k y^{n-k}$.

However, an illuminating combinatorial interpretation of the $q$-binomial coefficients is that they count points in a [[Grassmannian]] consisting of $k$-dimensional linear subspaces of an $n$-dimensional [[vector space]] $\mathbb{F}_q^n$ over a [[finite field]] $\mathbb{F}_q$ of cardinality $q$. This interpretation can be used to provide a bijective proof of the $q$-Pascal identity

\begin{displaymath}
\binom{n}{k}_q = \binom{n-1}{k}_q + q^{n-k}\binom{n-1}{k-1}_q.
\end{displaymath}
For, if we choose a standard affine hyperplane $H = \{x_n = 1\}$ in $\mathbb{F}_q^n$, then any linear $k$-plane $W$ in $\mathbb{F}_q^n$ either intersects $H$ in an affine $(k-1)$-plane in the affine $(n-1)$-space, or intersects $H$ in the empty set, i.e., $W$ is contained in the linear space $\{x_n = 0\}$. The number of $W$ satisfying the latter condition is $\binom{n-1}{k}_q$, and the number of $W$ satisfying the former is in bijection with the number of $(k-1)$-dimensional affine subspaces of $\mathbb{F}_q^{n-1}$, which is $q^{n-1}$ times the number of linear subspaces $\binom{n-1}{k-1}_q$ divided by $q^{k-1}$, the number of translations which stabilize a given affine subspace. (Maybe to be rewritten later.)

\hypertarget{counting_trees_cayleys_theorem}{}\subsubsection*{{Counting trees: Cayley's theorem}}\label{counting_trees_cayleys_theorem}

Cayley's theorem, here not to be confused with Cayley's observation that every group (monoid) is a subgroup (submonoid) of a permutation group (endofunction monoid), says that the number of tree structures on an $n$-element set $\{x_1, \ldots, x_n\}$ is $n^{n-2}$.

There are many proofs of this fact in the literature. One, due to Gilbert Labelle and made popular by Andr\'e{} Joyal, proceeds by establishing a bijection between trees equipped with a pair $(x_i, x_j)$ of nodes (allowing $x_i = x_j$) and the number of endofunctions on $X = \{x_1, \ldots, x_n\}$.

On the one hand, such a structure $(T, x_i, x_j)$ on $X$ can be identified with a linearly ordered set (the path $P$ from $x_i$ to $x_j$ in $T$) together with, for each node $y$ along that path, a rooted tree structure $T_y$ consisting of nodes $x \in X$ for which the shortest path in $T$ from $x$ to $P$ ends at $y$. (As a subgraph of $T$, this naturally forms a tree $T_y$ rooted at $y$.) Thus, such a structure $(T, x_i, x_j)$ on $X$ determines and is determined by an equivalence relation $E$ on $X$, a rooted tree structure on each equivalence class, and a linear ordering of the set $X/E$ of equivalence classes.

On the other hand, each endofunction on $X$, say $f: X \to X$, has an [[eventual image]] or stable set $stab(f) = \bigcap_{n \geq 0} f^n(X)$, on which $f$ acts by a permutation (i.e., bijectively). For each $y \in stab(f)$, there is a rooted tree $T_y$ which consists of all $x \in X$ for which the set of iterates $x, f(x), f^2(x), \ldots$ first lands in $stab(f)$ at $y$. Thus an endofunction on $X$ determines and is determined by an equivalence relation $E$ on $X$, a rooted tree structure on each equivalence class, and a permutation on the set $X/E$ of equivalence classes.

Since the number of permutations on $X/E$ is equinumerous with the number of linear orderings of $X/E$, we conclude that the set of ``bipointed tree'' structures on $X$ is equinumerous with the set of endofunctions on $X$. Thus the number of bipointed tree structures $(T, x_i, x_j)$ on $X$ is $n^n$, and therefore the number of tree structures $T$ on $X$ is $n^{n-2}$.

\hypertarget{vandermonde_identity}{}\subsubsection*{{Vandermonde identity}}\label{vandermonde_identity}

\hypertarget{stirling_number_identities}{}\subsubsection*{{Stirling number identities}}\label{stirling_number_identities}

\hypertarget{polynomial_identities}{}\subsection*{{Polynomial identities}}\label{polynomial_identities}

One method for proving [[polynomial]] identities (equations $p = q$ between polynomials $p, q \in \mathbb{C}[x_1, \ldots, x_k]$ in several variables) is first to give a bijective proof for the special case in which natural numbers $m_i$ are substituted for the variables $x_i$, and then reason that there are enough natural numbers that the polynomial identity must hold generally.

More formally, two polynomials are equal, $p(x_1, \ldots, x_k) = q(x_1, \ldots, x_k)$, if $p(m_1, \ldots, m_k) = q(m_1, \ldots, m_k)$ for all choices of natural numbers $m_i$. Relatedly, the set $\mathbb{N}^k \subseteq \mathbb{C}^k$ is Zariski-dense (i.e., the closure of $\mathbb{N}^k$ in $\mathbb{C}^k$ equipped with the [[Zariski topology]] is all of $\mathbb{C}^k$).

This is intuitively obvious of course. As a public service, we prove a more general result.

\begin{prop}
\label{}\hypertarget{}{}
Suppose $B$ is an [[infinite set|infinite]] [[subset]] of an [[integral domain]] $A$. Then a polynomial $p(x_1, \ldots, x_k) \in A[x_1, \ldots, x_k]$ is uniquely determined by its values it takes at arguments $b_1, \ldots, b_k \in B$.

\end{prop}
\begin{proof}
It suffices to show that if $p(b_1, \ldots, b_k) = 0$ for every choice of $b_i \in B$, then $p = 0$. We prove this by induction on $k$. The case $k = 0$ is trivial.

Assuming the assertion as induction hypothesis for case $k$, suppose $p \in A[x_1, \ldots, x_k, x_{k+1}]$ satisfies $p(b_1, \ldots, b_k, b) = 0$ for all choices $b_i \in B$ and $b \in B$. Then by induction, for each fixed $b \in B$ the polynomial $p(x_1, \ldots, x_k, b) \in A[x_1, \ldots, x_k]$ is identically zero. Thus if $F$ is the [[field of fractions]] of $A[x_1, \ldots, x_k]$, and if we regard $p$ as an element of the [[Euclidean domain]] $F[x_{k+1}]$ under the obvious embedding

\begin{displaymath}
A[x_1, \ldots, x_k, x_{k+1}] \cong A[x_1, \ldots, x_k][x_{k+1}] \hookrightarrow F[x_{k+1}],
\end{displaymath}
then $p$ has infinitely many [[roots]] $b$ in $A$ and thus in $F$, forcing $p$ to be identically zero.

\end{proof}


\end{document}