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We work in a category where pullbacks exist. Given a map $f:Y\to Z$, there is a pull-back diagram $$\begin{array}& Y \times_Z Y & \stackrel{g_1}{\longrightarrow} & Y \\ \downarrow{g_2} & & \downarrow{f} \\ Y & \stackrel{f}{\longrightarrow} & Z \end{array}$$ (I am sorry for the awful formatting, is there a better solution?)

Of course, because of symmetry, we should have $g_1=g_2$. But category theory is a formal language, so I would like to have a formal proof of this, without hand-waving. I have not managed to find one so far, although I think that it should not be difficult... The farthest I've got is that there must be a unique map $\phi: Y \times_Z Y \to Y \times_Z Y$ such that $g_1 \phi = g_2 \phi = g_1$, because of the universal property of the pullback. Similarly, there is a unique $\psi$ such that $g_1\psi=g_2\psi=g_2$. But how do I continue from here?

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It needn't be true in general that $g_1 = g_2$. In case you want to read more about such things, the morphisms $g_1$ and $g_2$ are called a kernel pair for $f$.

In the category of sets, for example, we can take $$Y \times_Z Y = \{ (a, b) \in Y \times Y \mid f(a) = f(b) \}$$ And then the functions $g_1 : Y \times_Z Y \to Y$ and $g_2 : Y \times_Z Y \to Y$ are given by projection onto the first and second coordinates, respectively. Then typically $g_1 \ne g_2$.

There are some cases where we might have $g_1 = g_2$. For example, if $f$ is a monomorphism, then since $f \circ g_1 = f \circ g_2$ it follows that $g_1 = g_2$. In fact, $f$ is a monomorphism if and only if $(\mathrm{id}_Y, \mathrm{id}_Y)$ is a kernel pair for $f$.

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