You have paid the price of admission: a topological space is a set together with a family of open sets obeying three axioms, and a basis lets you generate the whole family from a manageable seed. That is a lot of machinery to build for a single space. The reward is that you almost never build from scratch again. Once you have \mathbb{R}, you get the plane, the circle, the cylinder, the torus, the Möbius band and the Klein bottle essentially for free — by three constructions that each turn a space (or two) you already trust into a new one.

This page is about those three factories. Restrict to a subset and you get the subspace topology. Multiply two spaces and you get the product topology. Crush points together — glue, fold, wrap — and you get the quotient topology. Each is defined by a single clean rule, and each is the only sensible rule once you insist that the obvious maps (inclusion, projection, gluing) stay continuous. Learn the three rules and you can name most of the surfaces in an undergraduate topology course on sight.

The subspace topology: openness is relative

Let A be any subset of a topological space (X, \tau). We want to make A a space in its own right, and there is really only one honest choice: a subset of A should count as open when it is "the open part of X that you can still see from inside A". Formally, the subspace (or relative) topology is

This really is a topology: intersecting the axioms of \tau with A preserves them, since A \cap (U_1 \cap U_2) = (A \cap U_1) \cap (A \cap U_2) and A \cap \bigcup_i U_i = \bigcup_i (A \cap U_i), while A \cap X = A and A \cap \varnothing = \varnothing supply the two required members. Equivalently — and this is the way to remember it — the subspace topology is the coarsest topology on A making the inclusion map \iota : A \hookrightarrow X continuous.

The headline subtlety, the one that trips up every beginner, is that "open in A" is not the same as "open in X". A set can be perfectly open as a subset of A while being nothing like open in the ambient space.

• The half-open interval that is open. Give A = [0, 1] the subspace topology from \mathbb{R}. Then [0, \tfrac{1}{2}) is open in A, because [0, \tfrac{1}{2}) = [0,1] \cap (-1, \tfrac{1}{2}) is the trace of an open interval of \mathbb{R}. Inside [0,1], the point 0 has no "left" to escape to, so it counts as an interior point. In \mathbb{R} the same set is of course not open.
• The integers become discrete. Give \mathbb{Z} \subseteq \mathbb{R} the subspace topology. Each singleton is open, because \{n\} = \mathbb{Z} \cap (n - \tfrac{1}{2}, n + \tfrac{1}{2}). A subset with all singletons open has every subset open: the subspace topology on \mathbb{Z} is the discrete topology.
• The circle carries its own topology. The unit circle S^1 = \{(x,y) : x^2 + y^2 = 1\} \subseteq \mathbb{R}^2 inherits a topology whose open sets are the open arcs (traces of open discs). This is the topology we will build the torus out of below.

The product topology: rectangles, and why not just "all products"

Given spaces X and Y, we want a topology on the Cartesian product X \times Y. The natural building blocks are open rectangles U \times V, with U open in X and V open in Y. These do not themselves form a topology (a union of two rectangles is usually an L-shape, not a rectangle), but they form a basis: the product topology is the collection of all unions of open rectangles.

For two factors this is exactly what you would guess, and it recovers the familiar cases. On \mathbb{R} \times \mathbb{R} the open rectangles (a,b) \times (c,d) generate precisely the standard topology of the plane \mathbb{R}^2 — every open disc is a union of tiny rectangles, and every rectangle is a union of tiny discs, so the two bases agree. The product of the two circles, S^1 \times S^1, is the torus — one circle chooses your longitude, the other your latitude; the basic open sets are little curved rectangles (\text{arc}) \times (\text{arc}) on its surface.

The interesting question is what "the right topology" even means, and here the two-factor case hides a choice that only becomes visible for infinitely many factors. Consider a product \prod_{i \in I} X_i. Two candidates present themselves:

• The box topology: basis is all products \prod_i U_i with every U_i open. No restriction — you may shrink every coordinate at once.
• The product topology: basis is products \prod_i U_i in which U_i = X_i for all but finitely many i. You may constrain only finitely many coordinates at a time; the rest are left wide open.

For a finite product the two coincide — "all but finitely many are the whole space" is no restriction when there are only finitely many indices — which is why the distinction is invisible for X \times Y. For an infinite product they genuinely differ, and the product topology (the coarser of the two) is the one everyone means by default.

Why should making projections continuous force our hand? Continuity of \pi_j requires \pi_j^{-1}(U_j) = X_1 \times \cdots \times U_j \times \cdots to be open — a rectangle constraining exactly one coordinate. Finite intersections of these constrain finitely many coordinates, and their unions are all you can build. That is the product topology on the nose: it contains precisely the open sets forced by "projections are continuous", and not one more.

Worked example: openness of a rectangle, and continuity of a projection

Claim. In \mathbb{R}^2 with the product topology, the open square (0,1) \times (0,1) is open, and the projection \pi_1(x,y) = x is continuous.

Openness. The set (0,1) \times (0,1) is a single basic open rectangle — a product of two open intervals — so it is open by definition of the basis. (More interestingly, the open disc x^2 + y^2 < 1 is also open: around each interior point p you can fit a small axis-aligned rectangle inside the disc, so the disc is a union of basic rectangles.)

Continuity of the projection. Take any open U \subseteq \mathbb{R}. Then \pi_1^{-1}(U) = U \times \mathbb{R}, which is a product of two open sets, hence a basic open set of the plane. A map whose preimages of open sets are open is continuous, so \pi_1 is continuous — and the identical argument gives \pi_2. This is the finite-dimensional shadow of the theorem above: the product topology exists precisely so that these preimages land on the nose in the basis.

The quotient topology: gluing, folding, wrapping

The third factory is the most vivid. Start with a space X and a surjection q : X \to Y onto a set Y — think of q as an instruction sheet for gluing, telling you which points of X get fused into the same point of Y. (Equivalently, start from an equivalence relation \sim on X and let Y = X/{\sim} be the set of classes, with q sending each point to its class.) The quotient topology declares

In words: a set downstairs is open exactly when the bundle of points that map into it is open upstairs. This is a topology because preimage commutes with unions and intersections, and it is — dually to the subspace case — the finest (most open sets) topology on Y making q continuous. It is exactly the right amount of openness: any more and q would stop being continuous.

The magic is that this bland-looking rule captures our physical intuition of bending paper and joining edges. The cleanest example: take X = [0, 1] and glue its two endpoints, 0 \sim 1, leaving every interior point alone. The quotient [0,1]/(0 \sim 1) is homeomorphic to the circle S^1 — the map t \mapsto (\cos 2\pi t, \sin 2\pi t) sends both endpoints to (1,0) and is otherwise injective, and it carries the quotient topology to the circle's topology precisely. Bending a strip of paper round and taping its ends: made rigorous.

The real workhorse is gluing the edges of a square [0,1] \times [0,1], where the direction in which you identify each pair of opposite edges decides the surface. The interactive figure above walks through the torus recipe; the table records what the other choices produce.

Notice how much these constructions reuse the other two factories: the square is a subspace of \mathbb{R}^2 (itself a product), and the surfaces are quotients of it. New spaces from old, all the way down.

Worked example: the quotient criterion tells you what is open on the circle

Take q : [0,1] \to S^1, q(t) = (\cos 2\pi t, \sin 2\pi t), so q(0) = q(1) = (1,0). Which subsets of the circle are open in the quotient topology?

A small arc away from the glued point — say the image of (\tfrac14, \tfrac12) — has preimage exactly the open interval (\tfrac14, \tfrac12), which is open in [0,1]. So the arc is open. No surprise there.

A small arc straddling the glued point (1,0) is the subtle one. Its preimage is two pieces near the two ends, [0, \varepsilon) \cup (1 - \varepsilon, 1]. This set is open in the subspace [0,1] (each half is a trace of an open interval, using the endpoint rule from the subspace section). Because the preimage is open, the straddling arc is open too — which is exactly what we need for (1,0) to be an ordinary, boundary-free point of the circle. The gluing has genuinely healed the two ends into one interior point, and the quotient criterion is what certifies it. Reading "open downstairs" as "open preimage upstairs" turns a picture of taping paper into a checkable statement.