Consider the following category $\mathcal{C}$:
• Objects are pairs of sets $(X′, X)$ with $X′ ⊆ X$.
• Morphisms $(X ′ , X ) \rightarrow (Y ′ , Y )$ are functions $f : X \rightarrow Y$ such that $f^{−1}(Y') = X′$ i.e. such that for all $x\in X,\; f(x)\in Y′\iff x\in X′$.
Show that if a map $(X′, X) \rightarrow (Y ′, Y )$ of $\mathcal{C}$ is monomorphic then its underlying function $f : X \rightarrow Y$ is injective.
I am looking for hints and suggestions as to where to begin with this problem. At the moment I am not even sure how to get started. Thanks.
Suppose $F:(X',X)\to (Y',Y)$ is a monomorphism and let $x,y\in X$ satisfy $f(x)=f(y)$. We have two options: either $f(x)=f(y)\in Y'$ or not.
Suppose $f(x)=f(y)\in Y'$. Then both $x$ and $y$ are in $X'$, by definition of the morphisms in $\mathcal{C}$.
Let $(Z',Z)$ be an object of $\mathcal{C}$. Define a function $g:Z\to X$ which sends all of $Z'$ to $x\in X'$. You can easily see that this gives a morphism $G:(Z',Z)\to (X',X)$. Similarly, define a function $h:Z\to X$ which sends all of $Z'$ to $y\in X'$. Again, you can easily see that this gives a morphism $H:(Z',Z)\to (X',X)$. Moreover, these morphisms satisfy $F\circ G=F\circ H$ in $\mathcal{C}$, which implies, since $F$ is supposed to be a monomorphism, that $G=H$. But this can only happen if $x=y$, so $f$ is injective.
Now try to see what happens when $f(x)=f(y)$ is not in $Y'$.