The diagonal $\Delta_X \subseteq X \times X$, the diagonal map $\delta_X$, and the pairing $\langle f, g \rangle$ 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

$\Delta_X \;:=\; \{\, z \in X \times X : z_0 = z_1 \,\} \;=\; \{\, (x,x) : x \in X \,\} ,$

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

$\delta_X : X \to X \times X, \qquad \delta_X(x) := (x,x) ,$

that is, the function sending $x$ to the constant function $2 \to X$ with value $x$. Its two components are $\pi_0 \circ \delta_X = \mathrm{id}_X$ and $\pi_1 \circ \delta_X = \mathrm{id}_X$, 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 $\Delta_X$, and it is injective, since $\delta_X(x) = \delta_X(x')$ forces $x = x'$ by reading the coordinate at $0$. Whether $\delta_X$ is an embedding onto $\Delta_X$ (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

$\langle f, g \rangle : Z \to X \times Y, \qquad \langle f, g \rangle(z) := (f(z), g(z)) .$

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 $\pi_0 \circ \langle f, g \rangle = f$ and $\pi_1 \circ \langle f, g \rangle = g$; 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

$\delta_X = \langle \mathrm{id}_X, \mathrm{id}_X \rangle ,$

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 $f, g : Z \to Y$,

$\langle f, g \rangle^{-1}[\Delta_Y] \;=\; \{\, z \in Z : f(z) = g(z) \,\} ,$

directly from the definitions above: $\langle f, g \rangle(z) \in \Delta_Y$ 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 $\Delta_X$ 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.