Continuity of Maps

The preimage definition of continuity is the load-bearing joint of essentially all of modern geometry and analysis. In Homeomorphisms, two spaces are declared "the same" topologically precisely when a continuous bijection has a continuous inverse — continuity in both directions is the entire content of the definition. In The Subspace Topology, continuity of an inclusion map and of restrictions of continuous functions are handled by exactly the preimage machinery proved here. In Connected Spaces and compactness theory, the single fact that continuous images of connected (or compact) sets are connected (or compact) — itself proved via the preimage definition — underlies the Intermediate Value Theorem and the Extreme Value Theorem in full generality, not just on $\mathbb{R}$. Beyond pure topology: in algebraic geometry, morphisms of varieties and schemes are required to be continuous with respect to the Zariski topology, and the preimage definition is the only version available since the Zariski topology is almost never metrizable. In functional analysis, continuity of linear operators between infinite-dimensional spaces (with weak, weak-*, or norm topologies) is checked via preimages of subbasic sets, since these topologies are frequently not first-countable and sequential characterizations can fail. In computer science and domain theory, continuous functions between partially ordered sets equipped with the Scott topology model computable processes, again via the preimage definition. In probability and statistics, measurable functions are defined by an entirely parallel preimage condition (preimages of measurable sets are measurable), a direct structural cousin of topological continuity, and continuous functions on probability spaces underlie weak convergence of measures. In physics, continuity of field configurations and of gauge transformations on manifolds is stated in the same preimage language whenever the underlying spaces are not naturally metric (e.g. moduli spaces of connections).

Ask the tutor: Can you show me why a continuous bijection need not have a continuous inverse, to motivate the extra work in defining homeomorphisms?

Think of the open sets of a topology as the "test regions" a space provides — the collection of sets that count as neighborhoods of their points. A map $f: X \to Y$ is continuous when it never smuggles a discontinuity into $X$ that wasn't already visible in $Y$'s open sets: whenever $f(x)$ lands inside some open region $V$ of $Y$, the points of $X$ near $x$ must also land inside $V$. The definition via preimages captures exactly this, but stated globally and all at once: $f^{-1}(U)$ being open says that the set of points sent into $U$ is itself an "acceptable neighborhood-shaped" set. The classical $\varepsilon$-$\delta$ picture is a special case where $X,Y$ are metric spaces and open sets are unions of open balls. Continuity at $x_0$ demands: for every $\varepsilon$-ball $V = B(f(x_0),\varepsilon)$ around the image, there is a $\delta$-ball $U = B(x_0,\delta)$ around $x_0$ that maps inside $V$. That is precisely the pointwise definition above, specialized to balls. The topological definition simply asks that this control exists relative to every open set, not just balls, and packages it into a single global statement about preimages instead of a point-by-point quantifier chase. A useful mental picture: draw $X$ and $Y$ side by side. An open set $U\subset Y$ is a "blob." Continuity says: whatever the shape of $U$, when you pull the blob back through $f$, the resulting region in $X$ has no sharp edges — it is open. If $f$ tears a continuum into disconnected pieces or collapses an open interval to a single point without care, this pullback openness typically fails.

Example 1: a piecewise map on $\mathbb{R}$ with the standard topology, checked directly by the preimage definition. Let $X = Y = \mathbb{R}$ with the usual (metric/order) topology. Define

Note the asymmetry: the two pieces are genuinely different functions ($x^2$ vs. the affine $3x-2$), glued at the single point $x=1$. We check continuity two ways. First, continuity away from $x=1$. On $(-\infty,1)$, $f$ agrees with the polynomial $x^2$, which is continuous everywhere on $\mathbb{R}$ by standard $\varepsilon$-$\delta$ arguments (or: polynomials are continuous by algebra of continuous functions, itself provable from the basis criterion applied to sums/products). Restricting a continuous function to an open subspace keeps it continuous at points of that open set (this follows because for $x_0<1$ we can always shrink any neighborhood to lie inside $(-\infty,1)$, so local behavior only sees the $x^2$ branch). Similarly on $(1,\infty)$, $f$ agrees with the continuous affine function $3x-2$. Second, continuity exactly at $x_0 = 1$, via criterion (5) directly. We have $f(1) = 3(1)-2 = 1$. Let $\varepsilon>0$. We must find $\delta>0$ such that $|x-1|<\delta \implies |f(x)-1|<\varepsilon$.

• For $x \geq 1$: $|f(x)-1| = |3x-2-1| = 3|x-1|$. This is $<\varepsilon$ whenever $|x-1| < \varepsilon/3$.
• For $x<1$: $|f(x)-1| = |x^2-1| = |x-1|\,|x+1|$. Restrict attention to $|x-1|<1$ (so $0<x<2$, hence $|x+1|<3$); then $|f(x)-1| < 3|x-1|$, which is $<\varepsilon$ whenever $|x-1|<\varepsilon/3$.

Taking $\delta = \min(1, \varepsilon/3)$ handles both cases simultaneously: if $|x-1|<\delta$ then $|f(x)-1|<\varepsilon$ regardless of which branch $x$ falls in. So $f$ is continuous at $1$, and combined with continuity elsewhere, $f$ is continuous on all of $\mathbb{R}$. The key point the mismatched pieces expose: continuity at a gluing point is not automatic from continuity of each piece separately — it required checking that the two branches agree in value at $x=1$ ($1^2 = 1 = 3(1)-2$, a numerical coincidence we should double check: $1^2=1$ and $3\cdot1-2=1$, yes they match) and that both one-sided limits equal that common value. Had the affine piece instead been $3x - 1$ (giving $f(1)=2 \ne 1 = \lim_{x\to 1^-} x^2$), the function would have a jump discontinuity at $x=1$, and $f^{-1}\big((1.5,2.5)\big)$ would fail to be open (it would contain $1$ but no interval around $1$ from the left, since points just left of $1$ map near $1$, outside $(1.5,2.5)$).

Example 2: checking continuity via preimages of a subbasis, on a map between non-metric spaces — the Zariski-flavored finite-complement topology. Let $X = \mathbb{R}$ with the topology $\mathcal{T}_X$ where nonempty open sets are exactly the sets whose complement is finite (the cofinite topology), together with $\varnothing$. Let $Y=\mathbb{R}$ with the standard topology $\mathcal{T}_Y$. This example is deliberately asymmetric: domain and codomain carry different topologies, so no metric $\varepsilon$-$\delta$ shortcut applies on the domain side. Consider $f: X \to Y$, $f(x) = x^3 - 3x$ (chosen non-monotonic and non-injective, e.g. $f(-2) = -8+6=-2$ and $f(1) = 1-3=-2$ also — check: $1^3 - 3(1) = 1-3=-2$; indeed $f(-2)=f(1)=-2$, a genuine non-injectivity to keep the example honest). Claim: $f$ is continuous $X\to Y$. Since $\mathcal{T}_Y$ is the standard topology on $\mathbb{R}$, generated by the basis of open intervals $(a,b)$, by the basis criterion (statement (4) in the theorem) it suffices to show $f^{-1}((a,b))$ is open in $\mathcal{T}_X$ for every interval $(a,b)$, i.e. that $X \setminus f^{-1}((a,b))$ is finite (or all of $X$, or $f^{-1}((a,b))=\varnothing$). Compute $X\setminus f^{-1}((a,b)) = f^{-1}(\mathbb{R}\setminus (a,b)) = f^{-1}\big((-\infty,a] \cup [b,\infty)\big) = \{x : f(x) \leq a\} \cup \{x: f(x)\geq b\}$. Since $f(x) = x^3-3x$ is a nonconstant cubic polynomial, for any real number $c$ the equation $f(x)=c$ has at most $3$ real solutions (a nonzero polynomial of degree $3$ has at most $3$ roots — a standard algebraic fact). Moreover $f$ is a continuous real function tending to $-\infty$ as $x\to-\infty$ and $+\infty$ as $x\to+\infty$ (odd-degree leading behavior), so by the Intermediate Value Theorem applied to $f(x) - a$ and $f(x)-b$, the sets $\{f(x) \le a\}$ and $\{f(x)\ge b\}$ are each unions of finitely many closed rays and intervals determined by the (at most 3) roots of $f(x)=a$ and $f(x)=b$ respectively — in particular each such set, when it is not all of $\mathbb{R}$, is a finite union of intervals bounded by finitely many points, but this alone doesn't immediately give finiteness. We need a sharper argument, since $\{f(x)\le a\}$ is typically an infinite set (a ray), not a finite one. So instead we verify directly that $X \setminus f^{-1}((a,b))$ is finite by checking it can only fail to be finite if $(a,b)$ misses the entire bounded "middle range" of $f$ improperly — let us redo this correctly. Correct approach: We must show $f^{-1}((a,b))$ is open in the cofinite topology, i.e. either empty or cofinite (complement finite). Equivalently, its complement $\{x : f(x) \notin (a,b)\}$ must be finite whenever $f^{-1}((a,b))$ is nonempty — but this is false for generic $(a,b)$: e.g. take $(a,b) = (-100, 100)$; then $f^{-1}((-100,100))$ is a large but bounded interval-ish set (since $f\to\pm\infty$), and its complement $\{|f(x)|\ge 100\}$ is two unbounded rays — infinite, not finite, and not cofinite either (it's not all of $X$ since $f^{-1}((-100,100))$ is nonempty). This shows $f$ as defined is not continuous from the cofinite topology to the standard topology — a valuable correction that itself illustrates the theorem's power: the basis criterion gives a quick, honest test, and here the test fails cleanly. Let us instead verify the theorem's mechanism on the map that does work in this pairing, confirming the general principle. Correct positive example: let $g: X\to Y$ (cofinite domain, standard codomain) be any non-constant polynomial map is not continuous by the computation above (the preimage of a bounded interval containing enough of the range is never cofinite). The map that is continuous from $(\mathbb{R},\text{cofinite})$ to $(\mathbb{R}, \text{standard})$ is exactly a constant map $h(x) \equiv c$: for any open $V\subseteq Y$, $h^{-1}(V)$ is either $\varnothing$ (if $c\notin V$) or all of $X$ (if $c \in V$), and both $\varnothing$ and $X$ are open in any topology on $X$. So $h$ is continuous, while non-constant continuous real polynomials fail to be continuous when the domain is given the coarser cofinite topology. This corrected example teaches the real lesson: continuity depends on both topologies simultaneously, and a formula that looks "smooth" under the standard topology can fail to be continuous the instant the domain's topology is changed to something coarser like the cofinite topology — because the basis criterion must be checked against the actual open sets of $X$, not against intervals that happen to look natural.