A question about domain of composition of functions.

Question

This is more about convention I guess, but suppose we have

$f(x)=\frac 1x$

$g(x)=\frac 1{x-1}$

Then domain of $g$ is clearly all real number except $1$. But what would be the domain of $f\circ g$?

If we say $f\circ g(x)= x-1$ , then it is defined for all real number, but if we consider

$$f\circ g(x)= \frac 1{\frac 1{x-1}}$$

Then it is not defined for $1$, we would be the convention in this case. Should we use the simplified version? Or is it not a big deal and I can choose either one of those definition?

Answer 1

A function $h : A \rightarrow B$ is a binary relation $\{ (x,y) \in A \times B \mid y = h(x) \}$.

Function composition is therefore simply relational composition for binary relations

$R_1 \circ R_2 = \{ (x,z) \mid \exists y. (x,y) \in R_1, (y,z) \in R_2 \}$

where $R_1$ and $R_2$ are functions.

The answer is therefore that $dom(f \circ g) = dom(f)$, since the "intermediate" $y$ mentioned in the definition of relational composition must exist for the composite function to be defined for a given argument.

Answer 2

You must make sure that each step of the composition is defined. The domain of $f \circ g$ consists of all $x$ satisfying two properties: first, $x$ is in the domain of $g$, and second, $g(x)$ is in the domain of $f$. In your example, you must exclude $x=1$ for $x$ to belong to the domain of $g$, but since $g(x)$ is never $0$ there is no additional restriction.

A way to write this properly is $$f \circ g(x) = x-1, \quad x \not= -1.$$

As an analogy, the actions "take one step forward" and "take one step backward" are not always inverses: what if you are standing on the edge of the roof of a tall building?