I'm struggling with proving the following:
Let $A$ be a set such that $|A| \ge 2$ and let $f: A \to A$ be a mapping on $A$ such that $f(g) = g(f)$ for all mappings $g: A \to A$. Prove that $f$ is the identity function of $A$
I know this proof isn't particularly difficult but I can't for the life of me come up with a decent proof for it.
Suppose $f$ is not the identity function. Then there exist an $a_0\in A$ so that $f(a_0)\ne a_0$. Let $f(a_0) = a_1 \ne a_0$.
Now for any mapping $g:A\to A$ we must have that $g(f(a_0))= f(g(a_0))$ so $g(a_1)$ must equal $f(g(a_0)$.
Well what if we decide to select $g$ so that $g(a_0)= a_0$. As $g$ can be any mapping we can do this. So we must have $g(a_1) = f(g(a_0)) = f(a_0) = a_1$.
Well, what's forcing us to do that? As $g$ can be any mapping we can have $g(a_0) = a_0$ and $g(a_1) = a_0$.
We have a contradiction......
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Okay... let's make this simpler.
Let $a\in A$ and let $f(a) = b$. Select a $g$ so that $g(a)=a$ and $g(b) = a$. (As $g$ can be any mapping we can certainly do this.)
Therefore $f(g(a)) = g(f(a))$. But $f(g(a))= f(a) = b$. And $g(f(a))= g(b) = a$. So $a = b$ and $f(a) = b = a$.
So $f$ is the identity.
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Okay. Let's make this even simpler.
Let $a\in A$ but any element. Let $g$ be the constant function $g(x)=a$ for all $x \in A$. We must have $g(f(a)) = f(g(a))$ but $g(anything) =a$ so $g(f(a)) = a$ and $g(a)=a$ so we have $f(a) = f(g(a)) = g(f(a)) = a$.
So $f(a) =a$ for all $a \in A$.
So $f$ is the identity function.
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To make it a little more sophisticated (but not as simple).
Consider the class of mappings so that for any $w\in A$, $g_w$ is the constant mapping $g_w(\color{green}x)=w$ for all $\color{green}x\in A$.
Then for any $a\in A$ we must have $g_a(\color{green}{f(a)}) = a$. But we must have, as a property of $f$ that $g_a(f(a)) = f(g_a(\color{green}a))=f(a)$. So $a= f(a)$ for all $a\in A$.