The diagonal ${\mathrm{\Delta }}_X\subseteq X\times X$, the diagonal map ${\delta }_X$, and the pairing $⟨f,g⟩$ of two maps

Let $(X,\mathcal{T})$ and $(Y,{\mathcal{T}}_Y)$ be topological spaces (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison). Throughout, $X\times Y$ is the binary product ${\prod }_{i<2}{X}_i$ with $X_0=X$ and $X_1=Y$ (The product set ${\prod }_{i\in I}{X}_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space), carrying the product topology; a point of it is a function $z$ on the von Neumann natural $2=\{0,1\}$, written $(z_0,z_1)$, and ${\pi }_0,{\pi }_1$ are the two projections.

The basis used throughout. For the index set $2$ the product basis and the box basis coincide, since a box ${\prod }_{i<2}{U}_i$ has all but finitely many factors unrestricted for the trivial reason that it has only two (The product set ${\prod }_{i\in I}{X}_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). So

$\{U\times V:U\in \mathcal{T},V\in {\mathcal{T}}_Y\}$

is a basis for the product topology on $X\times Y$, and every statement below that tests a basic open set tests a box of two open sets.

The diagonal. The diagonal of $X$ is

${\mathrm{\Delta }}_X:=\{z\in X\times X:z_0=z_1\}=\{(x,x):x\in X\},$

the second description being the first read through the definition of a point of the product as a function on $2$. It is a subset of $X\times X$ and is given the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace) whenever it is regarded as a space.

The diagonal map. The diagonal map of $X$ is

${\delta }_X:X\to X\times X,{\delta }_X(x):=(x,x),$

that is, the function sending $x$ to the constant function $2\to X$ with value $x$. Its two components are ${\pi }_0\circ {\delta }_X={\mathrm{i}\mathrm{d}}_X$ and ${\pi }_1\circ {\delta }_X={\mathrm{i}\mathrm{d}}_X$, and by claim 2 of A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice it is the unique function $X\to X\times X$ with those two components. The same claim makes it continuous (Continuity of a map of topological spaces at a point and globally), the identity being continuous. Its image is ${\mathrm{\Delta }}_X$, and it is injective, since ${\delta }_X(x)={\delta }_X(x^{\mathrm{\prime }})$ forces $x=x^{\mathrm{\prime }}$ by reading the coordinate at $0$. Whether ${\delta }_X$ is an embedding onto ${\mathrm{\Delta }}_X$ (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological) is not asserted here; it is the content of the next item.

The pairing of two maps. For functions $f:Z\to X$ and $g:Z\to Y$ on a common domain, the pairing is

$⟨f,g⟩:Z\to X\times Y,⟨f,g⟩(z):=(f(z),g(z))\mathrm{.}$

By claim 2 of A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice it is the unique function $Z\to X\times Y$ with ${\pi }_0\circ ⟨f,g⟩=f$ and ${\pi }_1\circ ⟨f,g⟩=g$; no hypothesis on $f$ and $g$ is needed for the pairing to be defined, and continuity of the pairing is exactly continuity of both components, which is again that claim. In this notation

${\delta }_X=⟨{\mathrm{i}\mathrm{d}}_X,{\mathrm{i}\mathrm{d}}_X⟩,$

so the diagonal map is a special case of the pairing and needs no separate treatment.

The preimage identity that every later proof uses. For $f,g:Z\to Y$,

$⟨f,g{⟩}^{-1}[{\mathrm{\Delta }}_Y]=\{z\in Z:f(z)=g(z)\},$

directly from the definitions above: $⟨f,g⟩(z)\in {\mathrm{\Delta }}_Y$ says that the function $(f(z),g(z))$ on $2$ takes the same value at $0$ and at $1$.

• The diagonal is a subset of a product, and the diagonal map is a function into it; they are different objects with the same name. The set ${\mathrm{\Delta }}_X$ records which pairs are repetitions, and the map ${\delta }_X$ produces the repetitions. Both are needed: the closedness criterion of this page is about the set, and the transport of properties from $X$ to its copy inside the square is about the map.

• Nothing here depends on a choice principle. The product $X\times X$ is a binary product, and a point of it is exhibited by naming its two coordinates; the nonemptiness of an arbitrary product, which is where choice enters (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, claim 4), is never invoked for a binary product with a named point.

• Why the box description is recorded at the top. The criterion proved on this page tests basic open sets of $X\times X$, and for the binary product there is no gap between the box topology and the product topology to worry about (The product set ${\prod }_{i\in I}{X}_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). No infinite product is formed anywhere on this page, so the distinction never becomes live here.