Asymmetry of kernel pairs The kernel pair of an arrow is defined as the pullback of $A\overset{f}{\rightarrow} B\overset{f}{\leftarrow} A$. This definitions seems to be very symmetric, and yet the pullback $(P,\alpha,\beta)$ satisfies $\alpha=\beta$ if and only if $f$ is a monomorphism. I just noticed that I find this strange, since both $\alpha$ and $\beta$ are the pullback of $f$ along itself.
Can anybody explain what's going on? Why would $\alpha$ and $\beta$ generally differ? The proof that they're equal iff $f$ is a mono is clear to me, it just doesn't settle with the symmetry of the definition..
 A: Think of the case when $B$ is the terminal object and $f$ is the unique map $A\to B$. Then the pullback $P$ is just the product $A\times A$, and the maps $\alpha$ and $\beta$ are the projections on the two components. Clearly these are different!
But in general ($B$ isn't necessarily terminal now) there's a natural automorphism $\sigma:P\to P$ which "swaps" $\alpha$ and $\beta$, in the sense that $\alpha\circ \sigma = \beta$ and $\beta\circ \sigma = \alpha$. You obtain $\sigma$ from the universal property of the pullback by mapping $P$ to the first copy of $A$ by $\beta$ and to the second copy of $A$ by $\alpha$. In the case of the cartesian product of sets, this is the map $(x,y)\mapsto (y,x)$. I would view this is a manifestation of the symmetry that you're looking for: $\alpha$ and $\beta$ are not equal, but they're isomorphic in the category of arrows to $A$.
Edit: You've asked in the comments how to see that $\sigma$ is an isomorphism. Well, we have $\alpha\circ \sigma\circ \sigma = \beta\circ \sigma = \alpha$ and $\beta\circ \sigma\circ \sigma = \alpha \circ \sigma = \beta$. But the universal property of the pullback tells us that $\text{id}_P$ is the unique map $f\colon P\to P$ such that $\alpha \circ f = \alpha$ and $\beta\circ f = \beta$. Hence $\sigma\circ \sigma = \text{id}_P$, and $\sigma$ is self-inverse.
