Here's a lemma:
Is the following proof correct?
Suppose the square is a pullback. Then for all objects $Z$ and arrows $\alpha,\beta: Z\to X$ such that $f\alpha=f\beta$, there is a unique $\Gamma:Z\to X$ such that $\Gamma=\alpha=\beta$. In other words, whenever $f\alpha=f\beta$, we also have $\alpha=\beta$ (which is implied by the existence of $\Gamma$). Is that right? (In particular, is it right that we only need the existence of $\Gamma$, and its uniqueness is not used?)
Conversely, suppose that for all objects $Z$ and all arrows $\alpha,\beta:Z\to X$, $f\alpha=f\beta $ implies $\alpha=\beta$. We need to prove that there is a unique $\Gamma:Z\to X$ such that $\Gamma=\alpha=\beta$. So define $\Gamma$ to be $\alpha$ (which happens to be the same as $\beta$ when $f\alpha=f\beta$). Since the requirement $\Gamma=\alpha=\beta$, $\Gamma$ is forced to be $\alpha$, so it's unique. Is that right?