Proof: Composition of Continuous Maps is Continuous

The fact that a composition of continuous maps is continuous is one of the most basic yet important results in topology. It establishes that topological spaces and continuous maps form a category, and it is used implicitly in virtually every argument involving continuous functions.

Statement

Theorem2.12Composition of Continuous Maps

Let $f: X \to Y$ and $g: Y \to Z$ be continuous maps between topological spaces. Then the composition $g \circ f: X \to Z$ is continuous.

Proof via Open Sets

Proof

Let $W \subseteq Z$ be an open set. We must show that $(g \circ f)^{-1}(W)$ is open in $X$ .

By the definition of preimage under composition: $(g \circ f)^{-1}(W) = f^{-1}(g^{-1}(W)).$

Since $g$ is continuous and $W$ is open in $Z$ , the set $g^{-1}(W)$ is open in $Y$ .

Since $f$ is continuous and $g^{-1}(W)$ is open in $Y$ , the set $f^{-1}(g^{-1}(W))$ is open in $X$ .

Therefore $(g \circ f)^{-1}(W)$ is open in $X$ , and $g \circ f$ is continuous.

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Alternative Proof via Closed Sets

Proof

Let $C \subseteq Z$ be a closed set. Then $(g \circ f)^{-1}(C) = f^{-1}(g^{-1}(C))$ . Since $g$ is continuous, $g^{-1}(C)$ is closed in $Y$ . Since $f$ is continuous, $f^{-1}(g^{-1}(C))$ is closed in $X$ . Thus $g \circ f$ maps closed sets to closed preimages, hence is continuous.

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Alternative Proof via Closure

Proof

We use the characterization that $f$ is continuous if and only if $f(\overline{A}) \subseteq \overline{f(A)}$ for all $A$ .

Let $A \subseteq X$ . We need $(g \circ f)(\overline{A}) \subseteq \overline{(g \circ f)(A)}$ .

Since $f$ is continuous: $f(\overline{A}) \subseteq \overline{f(A)}$ .

Since $g$ is continuous: $g(\overline{B}) \subseteq \overline{g(B)}$ for any $B \subseteq Y$ . Taking $B = f(A)$ :

$g(f(\overline{A})) \subseteq g(\overline{f(A)}) \subseteq \overline{g(f(A))} = \overline{(g \circ f)(A)}.$

Thus $(g \circ f)(\overline{A}) \subseteq \overline{(g \circ f)(A)}$ , and $g \circ f$ is continuous.

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The Category of Topological Spaces

Remark$\mathbf{Top}$ as a Category

The composition theorem, together with the fact that identity maps are continuous, establishes that topological spaces and continuous maps form a category, denoted $\mathbf{Top}$ :

Objects: Topological spaces $(X, \tau)$ .
Morphisms: Continuous functions $f: X \to Y$ .
Composition: The composition $g \circ f$ as defined above.
Identity: The identity map $\operatorname{id}_X: X \to X$ .

The category axioms are satisfied:

Associativity: $(h \circ g) \circ f = h \circ (g \circ f)$ (this is a property of functions).
Identity law: $f \circ \operatorname{id}_X = f = \operatorname{id}_Y \circ f$ .

Isomorphisms in $\mathbf{Top}$ are precisely the homeomorphisms.

Generalizations

ExampleFinite Compositions

By induction, if $f_1: X_1 \to X_2, f_2: X_2 \to X_3, \ldots, f_n: X_n \to X_{n+1}$ are continuous, then $f_n \circ \cdots \circ f_1: X_1 \to X_{n+1}$ is continuous.

This is used repeatedly in analysis: for example, if $f: \mathbb{R}^n \to \mathbb{R}^m$ is continuous and $g: \mathbb{R}^m \to \mathbb{R}^k$ is continuous, then $g \circ f$ is continuous.

Theorem2.13Composition Preserves Open/Closed Maps

If $f: X \to Y$ and $g: Y \to Z$ are both open maps, then $g \circ f$ is an open map. Similarly for closed maps.

Proof

Let $U$ be open in $X$ . Since $f$ is open, $f(U)$ is open in $Y$ . Since $g$ is open, $g(f(U))$ is open in $Z$ . Thus $(g \circ f)(U) = g(f(U))$ is open. The argument for closed maps is identical.

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ExampleComposition of Homeomorphisms

If $f: X \to Y$ and $g: Y \to Z$ are homeomorphisms, then $g \circ f: X \to Z$ is a homeomorphism. Indeed:

$g \circ f$ is a bijection (composition of bijections).
$g \circ f$ is continuous (composition of continuous maps).
$(g \circ f)^{-1} = f^{-1} \circ g^{-1}$ is continuous (composition of continuous maps).

This proves that the homeomorphism relation is transitive.

RemarkContinuity and Restrictions

If $f: X \to Y$ is continuous and $A \subseteq X$ , then the restriction $f|_A: A \to Y$ is continuous (with the subspace topology on $A$ ), since $f|_A = f \circ \iota$ where $\iota: A \hookrightarrow X$ is the continuous inclusion.

Similarly, if $f(X) \subseteq B \subseteq Y$ , then $f: X \to B$ (viewed as a map to $B$ with its subspace topology) is continuous. The proof uses the fact that open sets in $B$ are of the form $V \cap B$ with $V$ open in $Y$ , and $f^{-1}(V \cap B) = f^{-1}(V)$ when $f(X) \subseteq B$ .