How to prove if something is a function?

Question

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

1. If $f: A \to B$ then the domain of the function should be A.
2. If ($z,x$) , ($z,y$) $\in f$ then $x = y$.

Now for example I have two functions:

1. $f:Z \to Z$
2. $g: Z \to Z$

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

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

1. 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.

2. 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.

3. 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.

Answer 1

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.