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I have the following writing problem: Let's say that $f,g\colon A \rightarrow B$ are given functions and I want to proof that $h = (f,g) \colon A \times A \rightarrow B \times B$ is a function.

Formally, I have to take $(a,b),(a,c)\in h$ and show that $b = c$. But $(a,b) ,(a,c) \in (f,g)$, then what? How can I formally conclude that $h$ is a function? I can't find the argumment. I hope that this is not a silly question.

Thanks in advance.

HeMan
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2 Answers2

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Usually, one has to check the good definition of a given function, since $A \times A$ could have diferent representatives for the same element. (For example, if your domain is $\Bbb{Z_4}$, here, $\bar{0}=\bar{4}=\bar{8}$ where the $\textit{bar}$ indicates the class). Now what does good definition of a function mean? Well, that for a given element in your domain, there is a unique element in the image which corresponds to it. Mathematically speaking,

given $f:A \to B$, then $f$ is a well defined function if $a,a' \in A$ such that $a=a'$ then $f(a)=f(a')$.

So how can you check for good definition in this case?

Well you have to get $(a,b),(a',b') \in A \times A$ such that $(a,b)=(a',b')$ and verify that $h((a,b))=h((a',b'))$ which in this case is true, since both $f$ and $g$ are said to be functions.

Joaquin Liniado
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  • I feel like you're muddying the explanation by bringing up equivalence classes. The original question doesn't refer to them. Also, I think your version of the definition of function clouds the issue, because it makes proving the original statement seem tautologous. In fact, you can't write “$f(a)$” until you know $f$ is a function. – Matthew Leingang May 03 '17 at 12:55
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I take it you're using this as your definition of function:

A subset $h \subset X \times Y$ is a function if, whenever $(x,y)$ and $(x,y')$ are in $h$, then $y=y'$.

This problem is complicated by the multiple set products, but if you work through it carefully, it's just a definition chase. [“Just” is in the eye of the beholder. With work, this kind of argument becomes routine.] The relation $h$ between $X = A\times A$ and $Y = B\times B$ is defined by $$ h = \left\{((a_1,a_2),(b_1,b_2)) \mid (a_1,b_1)\in f,\ (a_2,b_2)\in g\right\} $$ or, if you want, $$ h = \left\{((a_1,a_2),(b_1,b_2)) \mid f(a_1) = b_1,\ g(a_2) = b_2)\right\} $$

Suppose that $(x,y)$ and $(x,y')$ are in $h$. Then $x = (a_1,a_2)$, $y=(b_1,b_2)$, and $y' = (b_1',b_2')$ for suitable elements of $A$ and $B$, respectively. Then \begin{align*} (x,y) \in h &\implies (a_1,b_1) \in f \text{ and } (a_2,b_2) \in g \\ (x,y') \in h &\implies (a_1,b_1') \in f \text{ and } (a_2,b_2') \in g \\ \end{align*} Since $f$ is a function, we know $b_1 = b_1'$. Since $g$ is a function, we know $b_2 = b_2'$. Therefore $y=y'$. Thus $h$ also satisfies the definition of function.

Matthew Leingang
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