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I am pretty stumped on this question. All I knew was the definition of a monoid consisted of 3 properties:

  1. a Op b = c | Performing some operation on 2 values = some value c
  2. Zero = neutral element, where Zero Op a = a
  3. (a Op b) Op c = a Op (b Op c) | associativity

How do these properties help me show that a set of all surjectives x->x form a monoid? I was told there was composition involved, but I just have no clue how to go about it.

Asaf Karagila
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btrballin
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  • The only thing you’re missing is the operation, and the operation is composition: given two surjections $f$ and $g$, $f\circ g$ is a function which takes $x$ to $f(g(x))$. That is, it applies $g$ to $x$, and then applies $f$ to that. Try and prove that this is a surjection and it satisfies the three properties, with the identity as the neutral element. – Bob Jones Dec 06 '17 at 07:33
  • "Zero" is an awful name for the identity element $\ddot\frown$ – Angina Seng Dec 06 '17 at 07:34
  • @lord-shark-the-unknown Unless you denote by $+$ the operation of the monoid. – J.-E. Pin Dec 07 '17 at 07:57

1 Answers1

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Guide:

The binary operation that you are looking for is the composition function.

You have to check the following:

  • If $f$ is a surjection and $g$ is a surjection, can you prove that $f \circ g$ is a sujrection?

  • Show that the identity map, $e$, is a surjection. Also, $e \circ f = f \circ e$.

  • Check that $$(f \circ g) \circ h = f \circ (g \circ h)$$

Siong Thye Goh
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