Let's say I have a set $A$ and $a$ $\in$ $A$. Suppose I have a set $N$ = {f $\in$ $A^A$| $f(a) = a$}. Now, I want to check if $N$ forms a monoid of functions on $A$. Now, by verifying associativity and closure, I have completed two tests. But, now I want to see if there exists an identity function in $N$.
Confusion: Based on the definition of the set, it seems to me like every function behaves as an identity function because no matter what $f$, $g$ I pick in $N$, they will map every element in A to the same element in A, which is the definition of identity function.
Is this right? Or is there something fundamentally wrong about the way I am thinking about monoids?
Any help would be appreciated.
You have chosen one point $a\in A$. Your set of functions (in English) is the set of functions $f:A\to A$ which fix the point $a$. Among these functions is $\operatorname{Id}_A$ which is defined by $\operatorname{Id}_A(x)=x$ for all $x\in A$ because $\operatorname{Id}_A(a)=a$.