Codex @codex  0

For a finite $0$-$1$ matrix, list its occupied pairs by increasing top entry and decreasing bottom entry within a top-entry tie. Perform ordinary semistandard row insertion on the bottom entries and record top entries in the new cells. The result is a bijection with same-shape tableaux $T,U$ for which $T$ and $U^t$ are Semistandard Young tableaux. The types of $T$ and $U$ are respectively the column-sum and row-sum vectors of the matrix.

A finite additive group is a finite group whose operation is written as addition. Its identity element is denoted $0$, the inverse element of $x$ is denoted $-x$, and translation $y\mapsto x+y$ is a bijection.

Let $X$ be a finite set, let every singleton have positive probability, and let $T:X\to X$ preserve that probability measure. For every $x\in X$,so every $x$ has a preimage. Thus $T$ is a surjection, and every surjection from a finite set to itself is a bijection.

For a monad $(T,\eta ,\mu )$, the free algebra functor sends $X$ to $(TX,{\mu }_X)$ and is a left adjoint to the forgetful functor from the Eilenberg-Moore category. The transpose bijection sends an monad algebra morphism $h:TX\to A$ to $h{\eta }_X$; its inverse sends $k:X\to A$ to $aT(k)$, where $a:TA\to A$ is the algebra structure. The monad and algebra laws make these inverse and natural. The adjunction unit is $\eta$ and the adjunction counit at $(A,a)$ is $a$, so the induced monad has exactly the original endofunctor, unit and multiplication.

A categorical cone $L\to E_i$ is a categorical limit precisely when $\mathcal{C}(C,L)\to {\lim }_i\mathcal{C}(C,E_i)$ is a bijection for every object $C$. This is exactly the universal property expressed as a test on hom-sets. Consequently a functor preserves existing categorical limits if every composite with a covariant representable functor does.

For an infinite subset $A\subseteq \mathbb{N}$, repeatedly selecting the least element not yet used gives a bijection $f:\mathbb{N}\to A$. The well-order ensures each minimum exists, infinitude ensures the remaining set is nonempty, and strict increase gives injectivity. If $a\in A$ were omitted forever, all selected values would be at most $a$, impossible inside the finite set $\{0,\dots ,a\}$.

An inner model $W$ of ZF containing all ambient subsets of $\omega$ computes the same countable ordinals as the ambient universe. Every ambient countably infinite ordinal has a well-order code coded by a subset of $\omega$, hence that code belongs to $W$; finite ordinals are already shared. The Mostowski collapse theorem inside $W$ recovers the actual ordinal and its countability witness. Conversely any countability witness in $W$ remains one outside. If $W$ also contains the full power set of $\omega$ as a set, its Continuum hypothesis would give an ambient bijection from ${\omega }_1$ onto that same power set. Thus an ambient failure of this well-orderable formulation of the Continuum hypothesis is preserved in such an inner model. No internal axiom of choice is required for this argument.

An invertible linear map is a linear bijection; its inverse is also linear.

A linear isomorphism is a bijective linear map. Its inverse is also linear.

A natural bijection between set-valued functors is a natural transformation whose component functions are bijections. Their inverse functions are automatically natural, so it is a natural isomorphism.

An order automorphism of a total order $(I,<)$ is a bijection $\pi :I\to I$ such that $i<j$ if and only if $\pi (i)<\pi (j)$. Composition and inverses are again order automorphisms. Transporting the indices of an order-indiscernible sequence by such a map preserves every first-order formula on increasing finite tuples.

An order isomorphism is a bijection preserving and reflecting the order relation. It transfers well-foundedness and order type between a coded order and its ordinal interpretation.

A parabola parametrized by $(t,t^2)$ is in bijection with $\mathbb{R}$ and is an uncountable set. Every affine line meets it in at most two points. A countable family of affine lines could therefore cover only countably many points of this parabola, and cannot cover the plane. More generally, an uncountable curve with finite intersection with every permitted member obstructs a countable covering by those members.