Topology: Continuous Maps

Continuity in Metric Spaces

Following the case of real analysis, let’s define continuous functions via the usual ε-δ definition.

Definition. Let (X, d) and (Y, d’) be two metric spaces. A function f : X → Y is said to be continuous at $a\in X$ if:

for each ε>0, there exists δ>0 such that whenever $x\in X$ satisfies d(x, a) < δ, we have d’(f(x), f(a)) < ε.

Equivalently, we can replace the condition “whenever … ” by the set-theoretic statement $f(N_d(a, \delta))\subseteq N_{d'}(f(a), \epsilon)$ which is often more convenient.

If f is continuous at every a, we just say that f is continuous.

Theorem. The function f : X → Y is continuous at a iff for any open subset $V\subseteq Y$ containing f(a), there’s an open subset $U\subseteq f^{-1}(V)$ containing a.

In particular, f is continuous if and only if for any open subset $V\subseteq Y$ , the pull-back $f^{-1}(V)\subseteq X$ is open.

Proof.

Suppose f is continuous at a. If V is an open subset of Y containing f(a), then $N_{d'}(f(a),\epsilon)\subseteq V$ for some ε>0. By continuity at a, there exists δ>0 such that $f(N_d(a,\delta))\subseteq N_{d'}(f(a),\epsilon)$ . Hence $N_d(a,\delta) \subseteq f^{-1}(V)$ is an open subset containing a.

For the converse, suppose the stated condition holds. Given ε>0, $V := N_{d'}(f(a),\epsilon)$ is an open subset of Y containing f(a), so there is an open $U\subseteq f^{-1}(V)$ containing a. Since U is open, there’s an open ball $N(a,\delta)\subseteq U$ . It thus follows that $f(N(a,\delta))\subseteq V$ .

For the second statement, suppose f is continuous and $V\subseteq Y$ is open. By what we just proved, each element a of $U:= f^{-1}(V)$ is contained in an open subset $U_a \subseteq X$ which is contained in U. Thus U is a union of open subsets and is hence open.

Conversely, suppose whenever $V\subseteq Y$ is open, so is $U:=f^{-1}(V)\subseteq X.$ Let a be any point in X. From what we just proved, f is continuous at a. ♦

The theorem tells us that continuity is fundamentally a statement on the underlying topologies. In other words, if f : (X, d) → (Y, d’) is a continuous map of metric spaces, then we can replace d or d’ by any topologically equivalent metric and it wouldn’t make any difference.

Next, we prove that the metric itself is continuous (considering this map is so fundamental, it’d be pretty weird if it weren’t).

Proposition. In a metric space (X, d), the map $d:X\times X\to\mathbf{R}$ is continuous.

Proof

First, recall that there’s no fixed way of defining a metric on X × X, so we pick one of the many topologically equivalent ones, say $d_1((x,y), (x',y')) = d(x,x') + d(y,y').$

Let $(x,y)\in X\times X$ . To show continuity at that point, suppose ε>0. Letting δ = ε/2, we see that whenever $d_1((x, y), (x',y')) < \delta$ , we have: d(x, x’) < δ and d(y, y’) < δ and thus the triangular inequality gives:

d(x, y) – d(x’, y’) ≤ d(y’, y) + d(x, x’) < 2δ = ε;
d(x’, y’) – d(x, y) ≤ d(y, y’) + d(x’, x) < 2δ = ε.

Hence, |d(x’, y’) – d(x, y)| ≤ ε. This shows d is continuous at (x, y). ♦

Note

In addition, we had already proved that the standard arithmetic operations on real numbers are continuous. Thus, addition and product $\mathbf{R}\times \mathbf{R}\to \mathbf{R}$ are continuous, as is reciprocal $\mathbf{R}^*\to\mathbf{R}^*.$

Continuity in Topology

Since the notion of continuity can be described by open sets, let’s generalise it to topological spaces.

Definition. Let X and Y be topological spaces. A function f : X → Y on the underlying sets is said to be continuous at $a\in X$ if:

for any open subset $V\subseteq Y$ containing f(a), there’s an open subset $U\subseteq f^{-1}(V)$ containing a.

In particular, f is continuous (at every point of X) if for any open $V\subseteq Y$ , the pullback $f^{-1}(V)$ is open in X.

The following properties of continuity are obvious for any topological spaces X, Y and Z.

The identity map id : X→ X is continuous.
If f : X → Y is continuous at x and g : Y → Z is continuous at f(x), then gf : X → Z is continuous at x.
If T and T’ are both topologies on X, then id : (X, T) → (X, T’) is continuous if and only if T is finer than T’.

From the third property, one sees that if f : X → Y is bijective and continuous, the inverse $f^{-1}$ may not be continuous. Specifically, if T is strictly finer than T’, then the identity map (X, T) → (X, T’) is continuous but (X, T’) → (X, T) is not. Recall that if f : X → Y is bijective, continuous and has a continuous inverse, then we say f is a homeomorphism, in which case we get a bijection between the collection of open subsets of X and that of Y.

To check that a map is continuous, we don’t have to look at every open subset of Y.

Proposition. Let f : X→ Y be a map of topological spaces and $S'\subseteq \mathbf{P}(Y)$ be a subbasis of Y. Then f is continuous iff for any $V\in S'$ , $f^{-1}(V) \subseteq X$ is open.

Proof.

The forward direction is obvious. The converse follows from the fact that the pullback preserves arbitrary intersection and union:

$f^{-1}(W_1 \cap W_2 \cap\ldots \cap W_k) = f^{-1}(W_1) \cap f^{-1}(W_2) \cap\ldots \cap f^{-1}(W_k),$

$f^{-1}(\cup V_i) = \cup_i f^{-1}(V_i).$ ♦

For example, to check that f : X → R is continuous, it only suffices to prove that $f^{-1}((a, \infty))$ and $f^{-1}((-\infty, b))$ are open in X for any real a, b.

Next, we wish to prove that the standard maps are continuous.

Proposition.

If $Y\subseteq X$ is a subspace of a topological space, then the inclusion map $i:Y\hookrightarrow X$ is continuous.

If X and Y are topological spaces, then the projection maps $p_1: X\times Y\to X$ and $p_2 : X\times Y\to Y$ are continuous.

If $X_i$ is a collection of topological spaces and $X=\coprod_i X_i$ is the disjoint union, then every inclusion map $\iota_i : X_i \to X$ is continuous.

Proof

For the first statement, each open $U\subseteq X$ pulls back to $i^{-1}(U) = U\cap Y$ which is open in Y by definition of subspace. For the second statement, each open $U\subseteq X$ pulls back to $p_1^{-1}(U) = U\times Y$ which is open in X × Y by definition of product space. For the last statement, each open subset $U =\coprod_i U_i$ of X pulls back to $\iota_i^{-1}(U) = U_i$ which is open in X_i. ♦

In retrospect, one could even define topological spaces specifically to satisfy the above properties. And one defines the topology “just enough” such that the properties are satisfied. This will be expounded in a separate article.

Finally, we wish to show that continuity of a function depends “locally” on small neighbourhoods.

Theorem. If $f:X \to Y$ is a function of topological spaces and $X=\cup_i U_i$ is a union of open subsets $U_i \subseteq X$ , then

f is continuous if and only if $f|_{U_i} : U_i \to Y$ is continuous for every i.

Proof.

The forward direction is easy: since the inclusion map $\iota_i:U_i \hookrightarrow X$ is continuous, so is the composition $f\circ\iota_i = f|_{U_i}$ . Conversely, suppose each $f|_{U_i}$ is continuous. Let V be an open subset of Y. Then

$f^{-1}(V) = \cup_i (f^{-1}(V) \cap U_i) = \cup_i (f|_{U_i})^{-1}(V).$

Now each $(f|_{U_i})^{-1}(V)$ is open in $U_i$ which is in turn open in X. Thus, $(f|_{U_i})^{-1}(V)$ is open in X. So $f^{-1}(V)$ , being a union of open subsets of X, is open in X. ♦

Exercises

Suppose f : X → Y is a continuous function of topological spaces and $X_1\subseteq X$ and $Y_1\subseteq Y$ are subspaces such that $f(X_1)\subseteq Y_1$ . Prove that the restriction $g := f|_{X_1} : X_1\to Y_1$ is also continuous.
Suppose f : X₁ → Y₁ and g: X₂ → Y₂ are continuous maps. Then the concatenation $h:X_1\times X_2 \to Y_1\times Y_2,$ which takes $(x_1, x_2)\mapsto (f(x_1), g(x_2))$ , is also continuous.
Let Y be a topological space written as a union of subspaces $Y = \cup_i X_i.$ Prove that we get a continuous map $\coprod_i X_i \to Y$ which takes an element $x_i \in X_i$ to the corresponding image in Y.

Answers (Highlight to read)

Let V₁ be an open subset of Y₁. So we have V₁ = V ∩ Y₁ for some open subset V of Y. Then g^-1(V₁) = f^-1(V) ∩ X₁; by continuity of f, f^-1(V) is open in X, so f^-1(V) ∩ X₁ is open in X₁.
Let U₁ and U₂ be open subsets of Y₁ and Y₂ respectively; thus U₁ × U₂ is a basic open subset of Y₁ × Y₂. Then h^-1(U₁ × U₂) = f^-1(U₁) × g^-1(U₂) which is open in X₁ × X₂ since f and g are continuous. Thus the result follows.
Let V be an open subset of Y. The inverse image in disjoint union of X_i is the disjoint union of U_i, where each U_i = X_i ∩ V is open in X_i by definition of subspace.