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A continuous map on a nonempty topological space is point-transitive if it has a dense forward orbit. This is one convention for topological transitivity, but on spaces with isolated points it must be distinguished from the open-set intersection definition. On a nonempty compact metric space, the positive-time open-set intersection property implies point transitivity: for each member $U_j$ of a countable basis of a topology, ${\bigcup }_{n\ge 1}{f}^{-n}(U_j)$ is open and dense, and the Baire category theorem supplies a point in their intersection. Conversely, a surjective point-transitive continuous map has every tail of its dense orbit dense, since $f^k(X)=X$ and continuity sends its orbit closure into the closure of that tail. The map $f(1/j)=1/(j+1)$, $f(0)=0$ on $\{0\}\cup \{1/j:j\ge 1\}$ is point-transitive but fails the open-set intersection property.

A Polish space is a topological space whose topology can be induced by a complete separable metric. The completeness requirement concerns some compatible metric; it need not hold for every metric inducing the topology. Euclidean spaces are examples.

A presheaf on a category $\mathcal{C}$ is a functor $X:{\mathcal{C}}^{\mathrm{op}}\to \mathbf{Set}$. A morphism $f:C\to D$ induces the restriction map $X(f):X(D)\to X(C)$, reversing direction. The indexing category need not be small. For the category of open subsets of a topological space, this specializes to a presheaf of sets on a topological space.

A presheaf of sets on a topological space $X$ is a contravariant functor from the inclusion-ordered category $\mathcal{O}(X)$ of open subsets to $\mathbf{Set}$. Its maps $F(U)\to F(V)$ for $V\subseteq U$ are restriction maps.

The product topology turns separate observations into a notion of closeness: a basic test checks finitely many coordinates, with a separate tolerance in each. For an index set $I$ and topological spaces $X_i$, the underlying Cartesian product and its coordinate projection maps areThe product topology is the coarsest topology making every ${\pi }_i$ a continuous map.^[ref]

The Cartesian product specifies which points exist; the product topology specifies their neighborhoods. Finite observations motivate its finite-coordinate basis for the product topology, while convergence in a product space and continuity into a product space explain why it is useful. The compactness result for this construction is Tychonoff's theorem.

A scheme is quasi-compact when its underlying topological space is quasi-compact: every open cover has a finite subcover. Since affine open subsets form a base, this is equivalent to possessing a finite affine open cover. Every affine scheme is quasi-compact.

The quotient topology turns an instruction to glue points together into a precise topological space. Start with a topological space $(X,{\tau }_X)$ and an equivalence relation $\sim$. The quotient set $X/\sim$ consists of equivalence classes, and its canonical projection of a quotient set isThe quotient topology is defined by the rule using preimages of open sets:Thus a subset of the quotient space is open if and only if its entire preimage is open in $X$. Each fiber $q^{-1}(\{[x]\})$ is the whole equivalence class $[x]$, so the preimage includes every point glued to any of its members.^[ref]

This is the finest topology making $q$ a continuous map: every topology with that property can contain only subsets whose preimages are open. It really is a topology, because preimages preserve the empty set, the whole set, arbitrary unions, and finite intersections. The gluing example below shows why checking a complete preimage matters, especially at the seam.

The quotient topology also makes continuous maps out of a glued space easy to construct. A continuous map $f:X\to Z$ descends to a unique continuous map $\bar{f}:X/\sim \to Z$ exactly when it is constant on every fiber of $q$, meaning $x\sim y\Rightarrow f(x)=f(y)$. Define $\bar{f}([x])=f(x)$; the equivalence class representative cannot affect the answer. For every open set $V\subseteq Z$,The right side is open, so the quotient topology makes ${\bar{f}}^{-1}(V)$ open. Conversely, if $f=\bar{f}\circ q$ for a continuous map $\bar{f}$, then composition of continuous functions makes $f$ continuous, and equal equivalence classes force equal values. This is the universal property of the quotient topology. ^[ref]

A surjective continuous map is not automatically a quotient map. On the same two-point set $D=\{a,b\}$, take the identity map from the discrete topology to the indiscrete topology. It is continuous and surjective, but $\{a\}$ is not open in the target even though its preimage is open in the domain. The quotient topology for this identity map would be the discrete topology. Thus continuity supplies only one direction of the rule using preimages of open sets; a quotient map requires both.

Every open subset of a product of two topological spaces is a rectangle exactly when at least one factor has the indiscrete topology. For necessity, if both factors have nonempty proper open sets $U⊊X$, $V⊊Y$, then $(U\times Y)\cup (X\times V)$ is open, projects onto both full factors, and omits every point of $(X\setminus U)\times (Y\setminus V)$. A rectangle with both full projections would be the whole product, a contradiction. For sufficiency, if $X$ is indiscrete, every basic open rectangle is $X\times V$ or empty; an arbitrary union is $X\times \bigcup V$. The same argument works when $Y$ is indiscrete. Empty factors cause no exception because their topology is indiscrete.

A topological space is regular when a point and a closed set not containing it have disjoint open neighborhoods. Equivalently, for every point $x$ in an open set $U$, there is an open $W$ with $x\in W\subseteq \overline{W}\subseteq U$. To obtain shrinking, separate $x$ from the closed complement of $U$; the closure of the point's neighborhood avoids the other neighborhood. Conversely use such a shrinking and the complement of its closure. This definition alone does not require $T_1$.

For open subsets $U,V$ of a topological space $Y$, the relative cup product givesCompatibility with the maps to absolute cohomology identifies its image with the ordinary cup product. More generally the construction requires an excisive pair of subsets. The open-set case can be proved using subdivision to compute on chains subordinate to the open cover.

A retraction of a topological space $X$ onto a subspace $A$ is a continuous map $r:X\to A$ whose restriction to $A$ is the identity.

A ringed space $(X,{\mathcal{O}}_X)$ is a topological space $X$ equipped with a sheaf of rings ${\mathcal{O}}_X$. A morphism consists of a continuous map $f:X\to Y$ and a compatible morphism $f^{-1}{\mathcal{O}}_Y\to {\mathcal{O}}_X$.

The underlying topological space of every scheme is a Kolmogorov space: any two distinct points are distinguished by an open subset containing just one. Choose an affine open subscheme containing the first point. If it omits the second, it already distinguishes them. Otherwise, the two points correspond to distinct prime ideals, and an element in one but not the other defines a principal open subscheme distinguishing them. Thus distinct points of a scheme have distinct closures.

A second-countable topological space has a countable collection of open sets such that every open set is a union of members of that collection. A space covered by countably many open manifold charts is second countable because each coordinate domain has a countable basis of rational balls.

A topological space is semilocally simply connected if every point has an open neighbourhood whose loops based at that point become null-homotopic in the whole space. The neighbourhood itself need not be simply connected. Together with path connectedness and local path connectedness, this condition gives the classical existence and classification of connected covering spaces by subgroups of the fundamental group.

A topological space is separable when it has a countable dense subset. The topology is part of the assertion: a dual may be separable in its weak-star topology and not in its norm topology. A compact metric space is separable by taking the union of finite nets at scales tending to zero.

A separation axiom is a condition on a topological space expressing how points or closed subsets can be distinguished by open sets. The Kolmogorov space condition distinguishes distinct points by some open set; the T1 space condition makes every point closed; the stronger Hausdorff space condition gives disjoint neighbourhoods to distinct points.

A sequentially closed set is one that contains every limit of its convergent sequences. This is weaker than being a closed set in a general topological space, but the two conditions agree in metric spaces.

A topological embedding $\gamma :S^1\to X$ of a circle into a topological space. On an oriented surface, an essential such curve is noncontractible; on a punctured surface, nonperipheral means that it does not surround only one puncture.

For a topological space $X$ and abelian group $A$, singular cohomology is the cohomology of the cochain complex $C^n(X;A)=\mathrm{Hom}(C_n(X;\mathbb{Z}),A)$, where $C_n$ is the singular chain group. Its differential is dual to the singular boundary. This theory has relative cohomology, long exact sequences and the Excision theorem. On spaces that are not locally contractible it can differ from Čech cohomology.