Composition of continuous functions theorem

If two functions $ f: X \to Y $ and $ g: Y \to Z $ are continuous, then their composition $ g \circ f: X \to Z $ is also continuous.

This theorem states that if you have two continuous functions, $ f $ and $ g $, where:

$ f: X \to Y $
$ g: Y \to Z $

then the composite function, which is the result of applying $ f $ followed by $ g $, i.e., $ g \circ f $, will also be continuous.

In simpler terms, applying $ f $ first and then $ g $ produces a new function that is still continuous.

A practical example
Proof

A practical example

Let's look at an example where we compose the functions $ g \circ f(x) $, with $ f $ being the inner function and $ g $ the outer one.

$$ f(x) = x^2 \ \text{on} \ \mathbb{R} $$

$$ g(y) = \frac{y}{2} \ \text{on} \ \mathbb{R} $$

Both functions are continuous on $ \mathbb{R} $.

We want to determine if the composition $ g \circ f(x) $ is continuous over all of $ \mathbb{R} $.

Let’s consider the open interval $ (-2, 2) \subset \mathbb{R} $ as an example.

The image of the interval $ (-2, 2) $ under $ f $ is the interval $ (0, 4) $.

The output of $ f $ becomes the input for $ g $, meaning the codomain of $ f $ serves as the domain of $ g $.

The image of $ (0, 4) $ under $ g $ is $ (0, 2) $, another open interval.

Therefore, the preimage of the open interval $ (-2, 2) $ under the composition $ g \circ f(x) = \frac{x^2}{2} $ is the open interval $ (0, 2) $.

Since the preimage of an open set is open, the composition $ g \circ f(x) = \frac{x^2}{2} $ meets the condition for continuity on the interval $ (-2, 2) $.

By applying this reasoning to any open set in $ \mathbb{R} $, the preimage of the composition $ g \circ f(x) $ is always open.

This shows that the composition of these functions is continuous.

Proof

Now, let’s formally prove the continuity of the composition $ g \circ f(x) $.

$ f: X \to Y $
$ g: Y \to Z $

Take an open set $ U $ in $ Z $ (which is the range of $ g $). The preimage of $ U $ under $ g \circ f $ (i.e., $ (g \circ f)^{-1}(U) $) must be open in $ X $.

By the continuity of $ g $, the preimage of $ U $ under $ g $ is open in $ Y $.

Then, by the continuity of $ f $, the preimage of this open set under $ f $ is open in $ X $.

As a result, $ (g \circ f)^{-1}(U) $ is open in $ X $, which proves that $ g \circ f $ is continuous.

This concludes the proof, as a function is continuous if the preimage of any open set is also open.