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With some friends I am currently reading and trying to understand Category Theory by Steve Awodey. As I am no trained mathematician, even simple issues can halt my progress. One occurred when I tried to work through exercise 1b (NB: I am not asking for help on the execise). There the graph of a function $f: A\rightarrow B$ is defined as: $$G(f)= \{(a,f(a)) \in A \times B\ |\ a \in A\}.$$ The same definition can be found on the german Wikipedia.

Now, my question is: Why is the explicit specification of $a \in A$ necessary? Isn't this already implicitly included in $(a,f(a)) \in A \times B$, since the elemtents of $A \times B$ are all ordered tuples $(a,b)$ of $a\in A$ and $b \in B$?

I note in passing, that other definitions found on this very website (MathSE1, MathSE2, MathSE3) and the english Wikipedia seem clear to me.

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    $\begingroup$ By the usual conventions of set-builder notation, the unnecessary notation is the $\in A\times B$ part, not the $\in A$ part. I guess they added $\in A\times B$ just to remind the readers where the elements end up. $\endgroup$ Commented 2 days ago
  • $\begingroup$ @coiso Thanks for clarifying! $\endgroup$ Commented yesterday

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... other definitions found on this very website ... and the english Wikipedia seem clear to me.

These definitions say that the graph of a function $f : A \to B$ is defined by $$G(f) = \{(a,f(a)) |\ a \in A\}.$$

This automatically implies that $G(f) \subset A \times B$, thus it is indeed unnecessary that Awodey defines $$G(f)= \{(a,f(a)) \in A \times B\ |\ a \in A\}.$$ I guess he wanted to emphasize already in the definition that $G(f) \subset A \times B$.

What is unnecessary here is the explicit requirement $(a, f(a)) \in A \times B$. However, the explicit specification $a \in A$ is necessary, and it occurs in the definitions which are clear to you. This is standard Set-builder notation.

Writing for example $$G(f) = \{(a, f(a))\}$$ would perhaps be understood in the intended sense, but does not make sense formally. As it stands, the set $\{(a, f(a))\}$ is a singleton with an unspecified $a$. One can guess that $a \in A$ because it occurs as an argument of $f$, but nothing is really clear here.

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  • $\begingroup$ Thanks for your answer! If I understand it correctly, the condition $a \in A$ ensures that all elements of the domain ($A$) are considered. If it is left out $a$ could be an element of any subset of $A$, which still satisfies $(a,f(a)) \in A \times B$ but it possibly restricts the domain. $\endgroup$ Commented yesterday
  • $\begingroup$ @Anchises Correct! $\endgroup$ Commented yesterday

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