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

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.