LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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$\delta_X$ is a topological embedding of $X$ onto $\Delta_X$ , and $\langle f, g \rangle$ is continuous whenever $f$ and $g$ are

Statement

Let $X$ , $Y$ and $Z$ be topological spaces, with $X \times Y$ and $X \times X$ carrying the product topology and $\Delta_X$ the subspace topology (The diagonal $\Delta_X \subseteq X \times X$ , the diagonal map $\delta_X$ , and the pairing $\langle f, g \rangle$ of two maps, 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, 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). Then:

The pairing is continuous exactly when both components are. For functions $f : Z \to X$ and $g : Z \to Y$ , the pairing $\langle f, g \rangle$ is continuous if and only if $f$ and $g$ are continuous (Continuity of a map of topological spaces at a point and globally).
The diagonal map is an embedding. $\delta_X : X \to X \times X$ is injective and continuous, its image is $\Delta_X$ , and the corestriction $\delta_X^{0} : X \to \Delta_X$ , $\delta_X^{0}(x) = (x,x)$ , is a homeomorphism. So $\delta_X$ is an embedding (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological) and $X \cong \Delta_X$ .
The inverse of $\delta_X^{0}$ is the restriction of the projection $\pi_0$ to $\Delta_X$ , and this restriction agrees with the restriction of $\pi_1$ .

Claim 2 is what licenses reading a property of $\Delta_X$ as a property of $X$ : being a topological property is exactly invariance under homeomorphism (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

Facts & Assumptions

Given: Topological spaces $X$ , $Y$ , $Z$ ; the products $X \times Y$ and $X \times X$ with the product topology; functions $f : Z \to X$ and $g : Z \to Y$ ; the diagonal $\Delta_X$ with the subspace topology; and the maps $\delta_X$ , $\langle f, g \rangle$ of The diagonal $\Delta_X \subseteq X \times X$ , the diagonal map $\delta_X$ , and the pairing $\langle f, g \rangle$ of two maps.

[A1]

$\delta_X(x) = (x,x)$ and $\langle f, g \rangle(z) = (f(z), g(z))$ ; $\Delta_X = \{\, z \in X \times X : z_0 = z_1 \,\}$ ; and $\pi_0 \circ \delta_X = \pi_1 \circ \delta_X = \mathrm{id}_X$ , $\pi_0 \circ \langle f, g \rangle = f$ , $\pi_1 \circ \langle f, g \rangle = g$ (The diagonal $\Delta_X \subseteq X \times X$ , the diagonal map $\delta_X$ , and the pairing $\langle f, g \rangle$ of two maps).

[L1]

A map $h$ into a product is continuous if and only if every component $\pi_i \circ h$ is continuous, and every projection is continuous (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, claims 1 and 2).

[L2]

The identity map of a space is continuous, since the preimage of an open set under it is that open set (Continuity of a map of topological spaces at a point and globally).

[L3]

For $S \subseteq W$ with the subspace topology, a function $g_0 : Z \to S$ is continuous if and only if $\iota \circ g_0 : Z \to W$ is continuous, $\iota$ being the inclusion; and the restriction of a continuous map $W \to Y$ to $S$ is continuous (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).

[L4]

A continuous bijection whose inverse is continuous is a homeomorphism, and a map that is injective and restricts to a homeomorphism onto its image with the subspace topology is an embedding (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

Proof

technique · direct

1.1

Suppose $f$ and $g$ are continuous; then the components $\pi_0 \circ \langle f, g \rangle = f$ and $\pi_1 \circ \langle f, g \rangle = g$ of $\langle f, g \rangle$ are continuous, so $\langle f, g \rangle$ is continuous.

A1L1

1.2

Suppose $\langle f, g \rangle$ is continuous; then its components $f$ and $g$ are continuous.

A1L1

1.3

$\delta_X$ is continuous, its two components both being $\mathrm{id}_X$ , which is continuous.

A1L1L2

1.4

$\delta_X$ is injective: if $\delta_X(x) = \delta_X(x')$ then reading the coordinate at $0$ gives $x = x'$ .

1.5

The image of $\delta_X$ is $\Delta_X$ : each $(x,x)$ lies in $\Delta_X$ , and each $z \in \Delta_X$ satisfies $z = (z_0, z_0) = \delta_X(z_0)$ .

1.6

The restriction $p := \pi_0|_{\Delta_X} : \Delta_X \to X$ is continuous, being the restriction of the continuous $\pi_0$ to a subspace; and $\pi_0|_{\Delta_X} = \pi_1|_{\Delta_X}$ , since $z_0 = z_1$ for $z \in \Delta_X$ .

A1L1L3

2.1

Steps 1.1 and 1.2 together are claim 1.

step 1.1step 1.2

2.2

The corestriction $\delta_X^{0} : X \to \Delta_X$ is continuous, since composing it with the inclusion $\Delta_X \to X \times X$ gives $\delta_X$ , which is continuous by step 1.3.

step 1.3L3

2.3

$\delta_X^{0}$ and $p$ are mutually inverse: $p(\delta_X^{0}(x)) = \pi_0(x,x) = x$ for $x \in X$ , and $\delta_X^{0}(p(z)) = (z_0, z_0) = (z_0, z_1) = z$ for $z \in \Delta_X$ , the middle equality holding because $z_0 = z_1$ .

step 1.5step 1.6A1

3.1

By steps 2.2, 2.3 and 1.6 the map $\delta_X^{0}$ is a continuous bijection with continuous inverse $p$ , hence a homeomorphism, and its inverse is $\pi_0|_{\Delta_X} = \pi_1|_{\Delta_X}$ ; this is claim 3 and, with steps 1.4 and 1.5, claim 2.

step 1.4step 1.5step 1.6step 2.2step 2.3L4

4.1

Claims 1, 2 and 3 are steps 2.1, 3.1 and 3.1 respectively, so the lemma is proved.

step 2.1step 3.1∎

Remarks

The projection restricted to the diagonal is the inverse, and that is why no openness argument is needed. A continuous bijection is in general not a homeomorphism (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological); here the candidate inverse is available for free as a restriction of a projection, so the homeomorphism is exhibited rather than deduced.
Nothing in claim 2 uses a separation hypothesis. Every space, Hausdorff or not, sits inside its own square as the homeomorphic copy $\Delta_X$ . What the Hausdorff condition decides is a different question, whether that copy is closed, and that is the next item.
Claim 1 is the binary case of the characteristic property and is stated separately only because it is used constantly. For a product of two factors the general statement 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 specialises to exactly the displayed equivalence, and the pairing notation is what makes the specialisation usable without re-indexing.

Depends on

Used by

For continuous f, g : Z → Y with Y Hausdorff the agreement set { z ∈ Z : f(z) = g(z) } is closed in Z Corollary

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 36 results over 12 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Product topology (Wikipedia) (standard reference, not scraped)
Embedding (Wikipedia) (standard reference, not scraped)
Diagonal embedding (PlanetMath) (standard reference, not scraped)

δX\delta_XδX​ is a topological embedding of XXX onto ΔX\Delta_XΔX​, and ⟨f,g⟩\langle f, g \rangle⟨f,g⟩ is continuous whenever fff and ggg are