# What is $E^E$ in the specification of monoid $(E^E, \circ )$ of the functions from set $E$ to itself?

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.

• 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! Apr 23, 2021 at 8:13
• I do not understand the notation $E^E$. If I had $E=\{0, 1\}$, e.g., what would be $E^E$? Apr 23, 2021 at 8:18
• 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 Apr 23, 2021 at 8:50

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.)

• Very clear and exhaustive, thank you! Apr 23, 2021 at 8:57