Re this puzzle:

> Suppose \\(X\\) is any monoid and \\(F : X \to X\\) is any map with \\(F(fg) = F(f) F(g) \\) for all \\(f,g \in X\\). Is \\(F(1)\\) a central idempotent?

It's taken me a bit of fiddling to come up with a counterexample to that one, but I think I've got one...

First let's note that \\(F(1)F(1) = F(1.1) = F(1)\\), so \\(F(1)\\) is certainly idempotent. And \\(F(1)F(g) = F(1.g) = F(g.1) = F(g)F(1)\\), so \\(F(1)\\) commutes with anything in the image of \\(F\\). But \\(F(1)\\) need not be central (ie commute with anything in \\(X\\)), as the following example shows.

Let \\(2 = \\{0, 1\\}\\) be a two-element set, and let \\(X\\) be the set of maps \\(2\rightarrow 2\\).

So \\(X\\) has four elements:

— the identity map

— the map switching round \\(0\\) and \\(1\\)

— the constant map \\(c_0\\) sending everything to \\(0\\)

— the constant map \\(c_1\\) sending everything to \\(1\\)

\\(X\\) is a monoid under composition, but it is not commutative: \\(c_0\circ c_1 \neq c_1 \circ c_0\\)

Now let \\(F : X \rightarrow X\\) be the constant map sending everything in \\(X\\) to \\(c_0\\).

Then we certainly have \\(F(x\circ y) = F(x)\circ F(y)\\), since both sides are always \\(c_0\\).

But as noted above, \\(c_0\\) is not central because it does not commute with \\(c_1\\).

> Suppose \\(X\\) is any monoid and \\(F : X \to X\\) is any map with \\(F(fg) = F(f) F(g) \\) for all \\(f,g \in X\\). Is \\(F(1)\\) a central idempotent?

It's taken me a bit of fiddling to come up with a counterexample to that one, but I think I've got one...

First let's note that \\(F(1)F(1) = F(1.1) = F(1)\\), so \\(F(1)\\) is certainly idempotent. And \\(F(1)F(g) = F(1.g) = F(g.1) = F(g)F(1)\\), so \\(F(1)\\) commutes with anything in the image of \\(F\\). But \\(F(1)\\) need not be central (ie commute with anything in \\(X\\)), as the following example shows.

Let \\(2 = \\{0, 1\\}\\) be a two-element set, and let \\(X\\) be the set of maps \\(2\rightarrow 2\\).

So \\(X\\) has four elements:

— the identity map

— the map switching round \\(0\\) and \\(1\\)

— the constant map \\(c_0\\) sending everything to \\(0\\)

— the constant map \\(c_1\\) sending everything to \\(1\\)

\\(X\\) is a monoid under composition, but it is not commutative: \\(c_0\circ c_1 \neq c_1 \circ c_0\\)

Now let \\(F : X \rightarrow X\\) be the constant map sending everything in \\(X\\) to \\(c_0\\).

Then we certainly have \\(F(x\circ y) = F(x)\circ F(y)\\), since both sides are always \\(c_0\\).

But as noted above, \\(c_0\\) is not central because it does not commute with \\(c_1\\).