# Question about the domain for composition of functions

I have an assignment question, and while I dont want to go into details here, working through this question made me realize I'm not sure how to formally discuss the composition of functions.

The situation given to me is I know certain properties about the composition of three functions $f \circ g \circ h$. So if all three functions $f, g$ and $h$ all map from $X$ to $X$, and we want to prove some property of $g$ when we know certain properties about the composition, how do we relate $g$ in the composition to $g$ by itself?

To perhaps explain this more clearly, it seems to me that $g$ in the composition is different from g by itself because $g$ in the composition maps from the range of $h$ to $X$, and if $g$ is considered independently it maps from all of $X$ to $X$. So then surely anything we prove about the composition is only relevant for $g$ from $\text{range}(h)$ to $X$, and not relevant for $g$ from $X$ to $X$.

Is it possible to relate them? What sort of statements relate the two? For instance, if the rank nullity theorem applies to one (assuming it is a linear function) would any statements proved with the theorem apply to the other?

Sorry about the very vague question, but my assignment has very strict plagiarism rules, and I wouldnt want to break any of them.

If anyone could explain, even with examples to help illustrate, it would be very much appreciated!

• "in the composition $g$ maps from the range of $h$ to $X$". I would say no to that. Also "in the composition" (whatever that means) function $g$ maps from $X$ to $X$. The range of $h$ is a subset of $X$ but it does not change $g$. – drhab Aug 26 '16 at 14:38
• but surely when talking about g in the composition of the three functions, the only inputs that g can accept are from the range of h? If that is not correct could you please explain in a little more detail? – Guest Aug 26 '16 at 14:41
• What you say in your comment is correct. I am only arguing that we cannot speak of different $g$ when it arises in a composition. Function $g$ is determined by its graph, wich will not be affected by function $h$. – drhab Aug 26 '16 at 14:43
• Maybe I am misunderstanding the notation here, but is g maps from X to X, then it maps all elements of X to at least some elements in X, but in the composition $g o h$ the only elements g can map from are those it receives from h, ie range(h) – Guest Aug 26 '16 at 14:44
• would you mind explaining? I think there is probably some technical detail here that is getting me a little confused? I thought a function is defined by the its domain, range, and the rule mapping one from the other, so if g in the composition has a different domain than g when considered independently, arn't they different functions? Thus anything you can prove for one doesnt necessarily apply to the other? – Guest Aug 26 '16 at 14:46

In set theory a function $g$ is by definition a set of ordered pairs that has the property: $$\langle a,b\rangle\in f\wedge\langle a,c\rangle\in f\implies b=c$$

The set $A:=\{a\mid\text{some }b\text{ exists with }\langle a,b\rangle\in f\}$ is completely determined by $g$ and is its domain.

The set $B:=\{b\mid\text{some }a\text{ exists with }\langle a,b\rangle\in f\}$ is completely determined by $g$ and is its range or its image.

If we say in this context that $f:A\to C$ is a function then $A$ denotes the domain, and secondly the range of $f$ is a subset of $C$. Set $C$ does not necessarily coincide with the range, and is called the codomain of $f$. Note that it is not determined by $f$. In the special case where codomain and range coincide we say that $f$ is surjective.

(There are fields in mathematics though where a function is looked at as a triple $\langle A,g,C\rangle$ where $A$ is the domain, $g$ is the graph of the function and $C$ is the codomain.)

If $f:A\to C$ is a function and $g:C\to D$ is function then $g\circ f:A\to D$ is a function. This with $\langle a,d\rangle\in g\circ f$ if and only if some $c$ exists with $\langle a,c\rangle\in f$ and $\langle c,d\rangle\in g$. The existence of $f$ in composition $g\circ f$ has no effect on $g$ whatsoever. It was, is and will be the same set of ordered pairs.

Your question is indeed very vague, but from the properties of a composed function you can actually retrieve some information about une of its component in general.

For instance, if you know that $f\circ g$ is surjective, you can prove that $f$ is surjective. Independantly of whatever $g$ may be.

I don't know about your exact question, but for instance if you know that $f\circ g \circ h$ is bijective you can easily prove that $g$ is bijective, without doing any other assumptions.

Moreover, you can always rely on other known properties of $g$. For instance, assume that you know that $g$ is continuous, that $h = 1_{\mathbb{Q}}$, ie $h(x) = \begin{cases} 1 \mbox{ if } x \in \mathbb Q \\ 0 \mbox{ else } \end{cases}$, and that $g\circ h = \begin{cases} x^2 \mbox{ if } x \in \mathbb Q \\ 0 \mbox{ else } \end{cases}$.

You can deduce that on $\mathbb Q$, $g(x) = x^2$. But you know that $g$ is continous (this is what you say by "we prove about the composition is only relevant for g from range(h) to X"). But then, $\mathbb Q$ being dense in $\mathbb R$ and $g$ being continuous, you conclude that $g(x) = x^2$ for all $x \in \mathbb R$.