TITLE: Why do we have "another" definition for the kernel?
QUESTION [17 upvotes]: Why does the definition $\ker(f)=\{(a,a')\in A\times A: f(a)=f(a')\}$ exist?  This definition is for any sort of algebraic system and any sort of function.  But which came first... this definition or the more familiar one?  I haven't seen this used anywhere outside of pure algebra, maybe there are applications in other fields, maybe in functional analysis?  But what's the use?  If a function is injective in this setting then $\ker(f)$ is the diagonal of $A$.  But does this add anything?  Are there isomorphism theorems that use this language?  Do we loose anything or gain anything, when speaking about a function's kernel in this manner?
As an example, I had an old homework assignment I found (that prompted me to ask this)

Let $f:A\to B$ be surjective.  Then a map $h:A\to C$ can be factored over $f$ (i.e. $h=g\circ f$) for some $g:B\to C$, if only if, $\ker(f)\subset \ker(h)$. If this is the case then the map $g$ is unique.

I don't need (or want) a solution to this exercise (I did it years back and posting it here would only take from future students), so please do not post a solution (this is just an example of where the 
"undegraduate" kernel doesn't work).

REPLY [2 votes]: Concerning applications to other fields, I just had a look at my own book which I am writing on differential geometry, and realised that I have a section on equivalence kernels in there already. The reason for it being there is that such equivalence kernels are used in the definition of general fiber bundles. The idea is that the total space $E$ of a fiber bundle is partitioned into individual fiber spaces $\pi^{-1}(\{p\})$ for $p\in B$ by a projection map $\pi:E\to B$. In this case, there is no algebraic sum or product operation which can translate between one fiber space and another, and there's no additive or multiplicative identity in $B$ which singles out a special "coset" in $E$ from which the others can be obtained by simple translation.
The "quotient" in the "first isomorphism theorem" in this case is the set of fiber spaces at each point of the fiber bundle, and the isomorphism is between the point space $B$ and the set $\{\pi^{-1}(\{p\});\,p\in B\}$ of fiber spaces. Then when you add topology on the total space, you get a quotient topological space structure, and when you add differentiable structure, you get a quotient differentiable manifold structure.
Like a lot of the modern very abstract algebra, you can live without the high abstractions, but abstraction often leads to insights, economy of thought, and possibly further applications that weren't otherwise apparent.