Claim: Consider a category, such that for any two objects $X,Y$ there exists an object $Z$ and epimorphisms $Z\twoheadrightarrow X$ and $Z\twoheadrightarrow Y$. We call such a $Z$ and the epis a common cover of $X$ and $Y$. Let [![enter image description here][1]][1] be a Cartesian square. We shall write $[p]$ for a generalized element with representative $p$. Let $[p],[p']$ be two generalised elements of $P$ such that $f^*([p])=f^*([p'])$ and $g^*([p])=g^*([p'])$. Then $[p]=[p']$. Proof: Replacing the domains of $p$ and $p'$ with a common cover, we can assume, that they have the same domain $D$. The equation $g^*(p)=g^*(p')$ means that we have two commutative diagrams [![enter image description here][2]][2] withe the same $eta$. We can replace $D$ by $D'$ and get a diagram [![enter image description here][3]][3] which commutes for $p$ and for $p'$ separately. In the same way we get an arrow $D\to B$ such that the diagram [![enter image description here][4]][4] which commutes for $p$ and for $p'$. But now the uniqueness in the definition of a Cartesian square yields $p=p'$. [1]: https://i.sstatic.net/vHFxh.png [2]: https://i.sstatic.net/fFfXU.png [3]: https://i.sstatic.net/4S7vD.jpg [4]: https://i.sstatic.net/FaLOn.jpg