I'm currently reading Goldblatt's Topoi - The categorial analysis of logic and at some point he lets one verification to the reader and I must admit that I'm stuck.
He's just introduced $\mathbf{M}$-sets where $\mathbf{M}$ is a monoid, and he's showing that the category of $\mathbf{M}$-sets and multiplication-preserving maps is a topos. He also introduced the concept of a left ideal and called $L_M$ the set of left ideals of $M$, and his claim is that $L_M$ together with the multiplication $m\cdot B = \{n\mid nm \in B\}$; and with the true arrow $true : \{0\} \to L_M$, $true(0) = M$ is a subobject classifier in this category.
Then he considers a monic arrow $k: X\to Y$ and argues that the character $\chi : Y\to L_M$ associated with this subobject is defined by $\chi(y) = \{n \mid n\cdot y \in X\}$.
This is where I'm stuck : I've managed to show that this does make the associated square a pullback square, and that any other arrow $g:Y\to L_M$ that makes it a pullback square has $g(x) = M$ for $x\in X$. I've also shown that it has $\chi(y) \subset g(y)$ for all $y\in Y$, but I can't prove that $\chi$ is the only one that works...
Any indications?
Suppose $m\in g(y)$. We then have $g(my)=mg(y)=\{n\mid nm\in g(y)\}$. But since $m\in g(y)$, $nm\in g(y)$ for all $n$, so $g(my)=M$. Since $g$ forms a pullback square, this implies $my\in X$, so $m\in\chi(Y)$.