How to prove if something is a function? I know two conditions to prove if something is a function:


*

*If $f: A \to B$ then the domain of the function should be A.

*If ($z,x$) , ($z,y$) $\in f$ then $x = y$.


Now for example I have two functions:


*

*$f:Z \to Z$ 

*$g: Z \to Z$


And I have to show that the following are also functions:


*

*$h: Z \to Z$ defined as $h(x) = f(g(x))$.

*$h: Z \to Z$ defined as $h(x) = f(x) + g(x)$.

*$h: Z \to Z$ defined as $h(x) = f(x) \times g(x)$.


Now in all these cases to I would have to show that the $Dom(h) = Z$. Now I show this by showing $Dom(h) \subset Z$ and then showing $Z \subset Dom(h)$. Hence through this I am able to show $Dom(h) = Z$.
Now to show this for the three functions:
1 (a). $Dom(h) \subset Z$: $Dom(h) = Dom(f(g(x))) = Dom (g(x)) = Z$ . (As this is an equality can I use this statement instead showing both sides as subsets of each other?)
1 (b). $Z \subset Dom(h)$: (I don't understand how I would show this side.)
2 (a), (b). $Dom(h) = Dom(f(x) + g(x)$ = $Dom(f(x) + Dom(g(x)) = Z + Z = Z$
3 (a), (b). $Dom(h) = Dom(f(x) + g(x)$ = $Dom(f(x) * Dom(g(x)) = Z$ (Can I do this instead of showing them subsets for each side. Plus here is it correct to say $Dom(f(x)*g(x)) = Z.$ What if the domains of the two functions were different?
Now for each function I have to show that each element in the domain only maps to one element in the co-domain. (How would I show this?)
Is this method correct: 


*

*Let $(z,x), (z,y) \in h$. Then $(z,x), (z,y) \in f(g(x))$ hence $\exists a,b : (z,a), (z,b) \in g$ AND $(a,x), (b,y) \in f$. So as $g$ is a function $a = b$, and then as $f$ is a function $x=y$ hence $h$ is a function.

*Let $(z,x), (z,y) \in h$. Then $(z,x), (z,y) \in f(x) + g(x)$ hence $\exists a,b,c,d : (z,a), (z,b) \in g$ AND $(z,c), (z,d) \in f$. ($x=a+c$, $y=b+d$). So as $g$ is a function $a = b$, and then as $f$ is a function $c=d$ hence $x=y$ so $h$ is a function.

*Let $(z,x), (z,y) \in h$. Then $(z,x), (z,y) \in f(x) \times g(x)$ hence $\exists a,b,c,d : (z,a), (z,b) \in g$ AND $(z,c), (z,d) \in f$. ($x=a*c$, $y=b*d$). So as $g$ is a function $a = b$, and then as $f$ is a function $c=d$ hence $x=y$ so $h$ is a function.
 A: I would understand the statement "$Z$ is the domain of $f$" to mean
that $\forall (a,b)\in f: a \in Z$
and $\forall a \in Z: \exists b: (a,b) \in f$.
The first of those two conditions says that 
$\mathrm{dom}(f) \subseteq Z$,
and the second condition says that $Z \subseteq \mathrm{dom}(f)$.
The notation $f\colon Z \to Z$ just means that $\mathrm{dom}(f) = Z$
(because there is a $Z$ on the left of the arrow)
and that the range of $f$ is a subset of $Z$
(because there is a $Z$ on the right of the arrow).
In other words, in addition to the fact that $\mathrm{dom}(f) = Z$,
the statement that $f\colon Z \to Z$ also implies
$\forall (a,b)\in f: b \in Z$.
(There is no implied statement of the form $\forall b\in Z: P$ because the range is
not necessarily equal to $Z$.)
With this interpretation of $f\colon Z \to Z$ and the kind of careful
reasoning you followed in your proofs that $h$ maps each element of its domain
to only one element of its range (those proofs look fine to me, by the way),
you can show that $\mathrm{dom}(h) = Z$ in each of your three subproblems.
