Continuity and Homeomorphism

In first-year analysis you learned to sweat over continuity with \varepsilon's and \delta's: a function is continuous at a if, however small a target tube of radius \varepsilon you demand around f(a), you can find a radius \delta around a whose image lands inside. It works beautifully on the real line — but it is soaked in distance. It talks about |x - a| < \delta and |f(x) - f(a)| < \varepsilon, and a topological space has no ruler at all — only a declared collection of open sets.

So we need a definition of continuity that never once mentions a number. Remarkably, one exists, it is shorter than the \varepsilon–\delta version, and on \mathbb{R} it says exactly the same thing. That single idea — continuity as the pulling-back of open sets — is the engine of all topology, and it powers the subject's central question: when are two spaces "really the same"? The answer is a homeomorphism, and it is why a topologist genuinely cannot tell a coffee mug from a doughnut.

The definition that forgot how to measure

Here is the whole thing. Let f : X \to Y be a map between topological spaces. Recall that the preimage (or inverse image) of a set V \subseteq Y is

f^{-1}(V) = \{\, x \in X : f(x) \in V \,\} \subseteq X,

the set of inputs that land in V. This is defined for any map — it does not require f to be invertible; the symbol f^{-1} here is an operation on sets, not a function.

A map f : X \to Y between topological spaces is continuous when either of these equivalent conditions holds:

Open form: for every open V \subseteq Y, the preimage f^{-1}(V) is open in X.
Closed form: for every closed C \subseteq Y, the preimage f^{-1}(C) is closed in X.

The two forms are the same statement seen through De Morgan's laws. Preimage commutes with complement — f^{-1}(Y \setminus V) = X \setminus f^{-1}(V) — and "closed" just means "complement of open". So if every open set pulls back to an open set, then complementing both sides shows every closed set pulls back to a closed set, and vice versa. You may check whichever is more convenient for the problem in front of you.

Note what the definition pointedly does not mention: no \varepsilon, no \delta, no distance, no limit, not even a point. Continuity has become a statement purely about how f interacts with the two topologies. That is exactly the abstraction we wanted.

A natural first guess is "continuous means f sends open sets to open sets" — the forward direction. It is wrong, and it is worth seeing why. Take the perfectly continuous f(x) = x^2 on \mathbb{R}. The open interval (-1, 1) maps onto [0, 1), which is not open (it contains its boundary point 0). So continuous maps can crush open sets flat.

Maps that send open sets to open sets have their own name — open maps — and they are a genuinely different, useful class. But continuity lives in the backward direction. The intuition: to guarantee "nearby inputs give nearby outputs", you fix a neighbourhood of the output first and ask which inputs are safe — and safety must be an open condition. Pulling back is the direction that respects the logic "for every target tolerance, some input neighbourhood works".

Why this recovers \varepsilon–\delta

The topological definition had better agree with the analysis one on \mathbb{R} (with its usual topology, where "open" means "a union of open intervals"). It does — and unwinding the equivalence is the best way to feel why the definition is the right one.

Suppose first that f : \mathbb{R} \to \mathbb{R} is continuous in the \varepsilon–\delta sense at every point. Let V be open and take any a \in f^{-1}(V), so f(a) \in V. Because V is open, some ball (f(a) - \varepsilon,\, f(a) + \varepsilon) \subseteq V. The \varepsilon–\delta condition then hands us a \delta > 0 with |x - a| < \delta \Rightarrow |f(x) - f(a)| < \varepsilon — i.e. the whole interval (a - \delta, a + \delta) maps into V, so it lies inside f^{-1}(V). Every point of f^{-1}(V) thus has a little interval around it inside the set: f^{-1}(V) is open.

The converse runs the film backwards. If preimages of open sets are always open, fix a and \varepsilon > 0. The ball V = (f(a) - \varepsilon,\, f(a) + \varepsilon) is open, so f^{-1}(V) is open and contains a; hence it contains some (a - \delta, a + \delta), and that \delta is exactly the one \varepsilon–\delta asked for. The \delta-ball was just "the room open sets leave you". The general definition simply replaces "ball" with "open set" throughout — which is why it survives the jump to metric spaces and beyond, where the very same argument works with d(x,a) in place of |x - a|.

Seeing it: open pulls back to open

The picture below is the one to keep. A continuous curve y = f(x) runs across the plane. On the output axis we mark an open band V = (c, d) — an open interval, endpoints excluded (drawn hollow). Tracing each edge of the band horizontally to the curve and then down to the input axis carves out f^{-1}(V): the set of x whose height f(x) falls inside the band. Because the endpoints of V are excluded, so are the endpoints of the preimage — the pulled-back set is open too. Play the steps to watch the band descend to the axis.

For a monotone curve like this one the preimage is a single interval, but the principle needs no monotonicity: if the curve dipped in and out of the band several times, the preimage would be a union of open intervals — still open, since any union of open sets is open. Continuity is precisely the promise that this never fails, for any open target.

A gallery of quick verdicts

The open-preimage test makes many maps continuous "by inspection" — often with an argument so short it fits in a sentence.

Constant maps are always continuous. If f(x) = y_0 for all x, then for any V \subseteq Y the preimage f^{-1}(V) is either all of X (if y_0 \in V) or the empty set (if not). Both are open in every topology, so f is continuous no matter which topologies sit on X and Y.
The identity is continuous (when the domain topology is at least as fine). \mathrm{id} : X \to X has \mathrm{id}^{-1}(V) = V, so it is continuous exactly when every open set of the target is open in the source. On a fixed space with one topology this is automatic — but "identity from finer to coarser is continuous, and not conversely" is the whole reason topologies are ordered by fineness.
Every map out of a discrete space is continuous. In the discrete topology on X, every subset is open. So whatever f^{-1}(V) turns out to be, it is a subset of X and therefore open. There is simply no way to fail the test — discreteness is continuity's paradise.
Every map into an indiscrete space is continuous. The indiscrete (trivial) topology on Y has only two open sets, \varnothing and Y. Their preimages are \varnothing and X — always open. So there is nothing to check: any function into a space with no interesting open sets is continuous.

These two extremes are worth remembering together: the discrete topology makes every map out of it continuous, and the indiscrete topology makes every map into it continuous. They are the two ends of the fineness scale, and they bracket every honest example in between.

Continuity is closed under composition

The single most-used fact about continuity is also the easiest to prove in this language. If f : X \to Y and g : Y \to Z are continuous, so is g \circ f : X \to Z.

Take any open W \subseteq Z. Preimage of a composite unwinds outermost first:

(g \circ f)^{-1}(W) = f^{-1}\big(g^{-1}(W)\big).

Now read it inside-out. Because g is continuous, g^{-1}(W) is open in Y. Because f is continuous, the preimage of that open set, f^{-1}\big(g^{-1}(W)\big), is open in X. So the preimage of every open W is open, and the composite is continuous. No \varepsilon, no chasing two \delta's through each other — just two applications of one definition. This is the payoff of the topological viewpoint: theorems that were fiddly in analysis become one-liners.

Homeomorphism: the topological "equals"

Every branch of mathematics has a notion of "the same": groups have isomorphisms, vector spaces have linear isomorphisms, metric spaces have isometries. Topology's version is the homeomorphism.

A map f : X \to Y is a homeomorphism when all three hold:

f is a bijection (one-to-one and onto);
f is continuous;
its inverse f^{-1} is continuous too.

When such an f exists, X and Y are homeomorphic, written X \cong Y.

A homeomorphism is a bijection between the points that is simultaneously a bijection between the open sets: U is open in X if and only if f(U) is open in Y. It is a perfect dictionary translating one space's topology into the other's, with no meaning lost either way. That is why homeomorphic spaces are, to a topologist, literally the same object wearing different clothes: any statement expressible purely in terms of open sets is true of one exactly when it is true of the other. Bending, stretching, and twisting are allowed; tearing (which rips open sets apart) and gluing (which fuses them) are not.

Worked homeomorphism: (0, 1) \cong \mathbb{R}. The open interval and the whole real line look wildly different — one is a short bounded segment, the other runs forever — yet topologically they are indistinguishable. A witness is

f : (0, 1) \to \mathbb{R}, \qquad f(x) = \tan\!\left(\pi x - \tfrac{\pi}{2}\right).

It is a continuous bijection (as x sweeps 0 \to 1, the argument sweeps -\tfrac{\pi}{2} \to \tfrac{\pi}{2} and \tan sweeps -\infty \to +\infty), and its inverse f^{-1}(y) = \tfrac{1}{\pi}\big(\arctan y + \tfrac{\pi}{2}\big) is continuous too. So "bounded" is not a topological property — it depends on the metric, which a homeomorphism is free to distort. A simpler cousin: any two open intervals (a, b) \cong (c, d) via the straight-line map h(x) = c + \frac{d - c}{b - a}(x - a).

The subject's most famous joke: to a topologist, a coffee mug and a doughnut are the same thing. Picture a mug made of infinitely stretchy clay. The cup part flattens down and merges into the handle; all the material migrates into a single ring. What survives every squash is the one feature you cannot remove without tearing: the single hole — the mug's through the handle, the doughnut's through the middle. Both are surfaces of genus 1 (one handle/hole), so there is a homeomorphism between them, built by exactly such a continuous deformation.

The same lens sorts the whole alphabet. The letters C, I, L, S, U, V, Z are all homeomorphic (each is a single unknotted arc). O, D form another class (a loop, one hole). A, R share a class (a loop with a tail or two — one hole). And X stands alone (four arms meeting at one special point). Topology sees past the font straight to the connectivity.

Topological invariants: proving spaces are different

Exhibiting a homeomorphism proves two spaces are the same. But how do you prove two spaces are not homeomorphic? You cannot inspect infinitely many candidate maps and reject each one. Instead you find a topological invariant: a property expressible in terms of open sets, hence automatically preserved by any homeomorphism. If X has the property and Y does not, no homeomorphism can exist — a single mismatch settles it.

The staple invariants:

Cardinality. A bijection is required, so homeomorphic spaces have the same number of points. (A blunt instrument, but it instantly rules out, say, a two-point space versus a three-point one.)
Connectedness. "Cannot be split into two disjoint non-empty open pieces" is a purely open-set condition, so it transfers across homeomorphisms — as does the finer count of connected components.
Compactness. "Every open cover has a finite subcover" mentions only open sets and is likewise preserved. A continuous image of a compact space is compact, so a homeomorphism carries compactness both ways.
The number of "holes". Made precise by algebraic topology (the fundamental group, homology), the count of independent loops you cannot contract is invariant — this is what distinguishes the mug/doughnut (one hole) from the sphere (none).

Three classic non-homeomorphisms, each cracked by one invariant:

[0, 1] \not\cong (0, 1). The closed interval is compact (closed and bounded, by Heine–Borel); the open interval is not (the cover by \big(\tfrac1n, 1 - \tfrac1n\big) has no finite subcover). Compactness differs, so no homeomorphism exists — even though there are plenty of continuous bijections nearby.
The circle S^1 \not\cong an interval. Use the remove-a-point trick: deleting any single point from an interval breaks it into two pieces (disconnected), while deleting any point from a circle leaves it in one piece (an arc is still connected). A homeomorphism restricts to a homeomorphism after removing matching points, so it would have to preserve "does removing a point disconnect the space?" — it does not match, so S^1 is not homeomorphic to any interval.
\mathbb{R} \not\cong \mathbb{R}^2. Same trick, one dimension up: remove a point from the line and it falls into two components; remove a point from the plane and it stays connected. Connectedness-after-puncture tells the line from the plane.

The most common error is to drop the third clause and declare "continuous and bijective \Rightarrow homeomorphism". It is false — the inverse can fail to be continuous, and then the map tears open sets even though it never merges points.

The canonical counterexample wraps a half-open interval around a circle:

f : [0, 2\pi) \to S^1, \qquad f(t) = (\cos t, \sin t).

This f is a continuous bijection: every angle in [0, 2\pi) hits a distinct point of the circle, and it is onto. But its inverse is not continuous. Near the point (1, 0) = f(0), points of the circle just below the positive x-axis correspond to parameter values near 2\pi — far from 0. So a set of circle points arbitrarily close to f(0) maps back to parameters split between "near 0" and "near 2\pi": the inverse rips a connected neighbourhood in two. Concretely, [0, \pi) is open in the domain [0, 2\pi), but its image is a half-open arc that is not open in S^1 — so f is not an open map, and f^{-1} is not continuous.

The invariants back this up: [0, 2\pi) is disconnected by removing an interior point, the circle is not — they were never homeomorphic to begin with, so no continuous bijection between them could have a continuous inverse. The lesson: on compact-to-Hausdorff maps a continuous bijection is automatically a homeomorphism (a genuinely useful theorem), but drop compactness and the inverse needs its own continuity check every single time.