Given any functions $f:X\to Y$ and $g:A\to B$ the function $h(x)=f(g(x))$ is well defined for any elements $x\in g^{-1}(X\cap g[A])$ can one then write $h=f\circ g$? Or is composition of $f$ and $g$ only defined when the domain of $g$ equals the codomain of $f$? If the composition is still well defined for some values, then why limit the definition? I understand that this could give rise to cases where you have functions with empty domains, in the circumstance the composition isn't defined anywhere but is that really a problem? Would it still be okay to write $h=f\circ g$?

  • $\begingroup$ When you write $h=f\circ g$, it is usually implicitly assumed that $g(A)\subset \text{dom}(f)$ $\endgroup$ – user160738 Jun 19 '17 at 2:40
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    $\begingroup$ To composite the functions $f:X\to Y$ and $g:A\to B$, the sufficient condition for $f\circ g$ to be defined on $A$ is that $B\subseteq X$. $\endgroup$ – BAI Jun 19 '17 at 2:43
  • $\begingroup$ @BAI But the function $h(x)=f(g(x))$ could still be well defined for values of $x$ even if this is not the case, I don't understand. Take $f:\mathbb{R}\to \mathbb{R}$ where $f(x)=x^2$ and $g:\mathbb{R}\to\mathbb{C}$ where $g(x)=x^3$ then $f(g(x))$ is still well defined for values of $x$ and yet $\mathbb{C}$ is not a subset of $\mathbb{R}$. $\endgroup$ – nomad66 Jun 19 '17 at 2:44
  • $\begingroup$ @nomad66 it seems like $x\in g^{-1}(X\cap g[A])$ implies that $x\in A$ and $f\circ g$ is defined for which $(x\in A)\land(g(x)\in X)$. You could just see it like we choose another pair of domain and codomain $g:A'\toB'$ such that $B'\subseteq X$. $\endgroup$ – BAI Jun 19 '17 at 2:54
  • $\begingroup$ @nomad66 and $B'\subseteq B$ $\endgroup$ – BAI Jun 19 '17 at 2:55

We need to work with one of two definitions:

Definition 1: Given two functions $f\colon X \to Y$ and $g\colon A \to B$, their composite $h = f\circ g$ is the function $h\colon g^{-1}(X \cap g[A]) \to Y$ given by $h(x) = f(g(x))$.

Definition 2 Given two functions $f\colon X \to Y$ and $g\colon A \to B$ such as that $g[A] \subseteq X$, their composite is the function $h\colon A \to Y$ given by $h(x) = f(g(x))$.

With both definitions, $h$ is well defined. Using definition 1, we can have an empty composite, a function from the empty set, while using definition 2, given that our sets are non-empty, then the composite is non-empty.

Also, when they apply, both definitions agree. For the sake of simplicity (given that we don't want to mess around with empty functions), we usually work with the second, less general definition.

Now, we may find ourselves in the middle way of the two extremes: we don't have $g[A] \subseteq X$, but their intersection is non-empty either: $X \cap g[A] \neq \varnothing$. For instance, take $g\colon \mathbb{R} \to \mathbb{R}$ and $f\colon \mathbb{R} - \{0\} \to \mathbb{R}$ given by $g(x) = x^2 - 1$ and $f(x) = \dfrac{1}{x}$. To define their composite $h$, we usually restrict $g$ to the pre-image of the non-problematic points, so we would be considering as it's domain the set $\mathbb{R} - \{\pm 1\}$ instead of simply $\mathbb{R}$. That's usually implicit when talking about composite functions, because when this happens, this restriction can always be done.


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