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\$$.