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In Wikipedia article about idempotence there is a specification of monoid $(E^E, \circ )$ of the functions from set $E$ to itself.

What is $E^E$ in this specification? Conceptually and, maybe, with some simple example, e.g. on a set of two elements.

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    $\begingroup$ It is the set of functions from $E$ to itself, as you said. Could you specify what is your question? I read the Wikipedia page but I can't see what you need. I don't want to be rough! Just help you better :) welcome on MSE! $\endgroup$ Apr 23, 2021 at 8:13
  • $\begingroup$ I do not understand the notation $E^E$. If I had $E=\{0, 1\}$, e.g., what would be $E^E$? $\endgroup$ Apr 23, 2021 at 8:18
  • $\begingroup$ I reckon that these are all possible total functions from $E$ to $E$, i.e. $\{\langle 1, 1 \rangle, \langle 2, 1 \rangle \}$ (both inputs mup to 1), and 3 other possible combinations, but I did not find the meaning of notation $A^B$ in general $\endgroup$ Apr 23, 2021 at 8:50

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Given two sets $A$ and $B$, we define $A^B$ to be the set of all functions from $B$ to $A$. In your particular example, we have $A = B = E$ and so the set can be given a monoid structure by defining the operation to be composition.

To answer your question from the comments: "What is $E^E$ when $E = \{0, 1\}$?"
Consider the following four functions $f_1, \ldots, f_4 : E \to E$

  1. The identity function, i.e., $f_1(x) = x$ for all $x \in E$,
  2. The constant $0$ function, i.e., $f_2(x) = 0$ for all $x \in E$,
  3. The constant $1$ function, i.e., $f_3(x) = 1$ for all $x \in E$,
  4. The switch function, i.e., $f_4(x) = \begin{cases}0 & x = 1 \\1 & x = 0\end{cases}.$

Then, $E^E = \{f_1, f_2, f_3, f_4\}$.

It would be a good exercise to verify which elements are idempotents. (There's only one which isn't.)

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    $\begingroup$ Very clear and exhaustive, thank you! $\endgroup$ Apr 23, 2021 at 8:57

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