# Why do we have "another" definition for the kernel?

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).

• Which is the "undergraduate" definition of kernel? I've only seen it defined for sets with some structure (so it seems to me your definition is a generalization of the definition I've seen). Aug 1, 2015 at 0:51
• Let $f:X\to Y$ such that $Y$ has identity called $y$, then $\ker(f)=f^{-1}(y)$. Aug 1, 2015 at 0:53
• – user14972
Aug 1, 2015 at 17:29

Fascinating question! I just checked my algebra books and functional analysis books, and found only one of the algebra books used the word "kernel" in the sense you mean. That was Mac Lane, Birkhoff, "Algebra", third edition, page 33. They call it an "equivalence kernel". This is clearly a generalization of the usual concept of a kernel as the inverse image of the identity. In an algebraic structure, you can specify an equivalence relation by specifying a subgroup, subring, submodule, etc., and then define the equivalence relation in terms of cosets etc. But in the absence of an algebraic structure, you can't use cosets. You have to specify the equivalence relation explicitly.

But there's a third definition of kernel, which is sort of related, in the functional analysis literature, namely an integration factor which resembles a Green's function in intergral equations. The 1950 Riesz/Sz-Nagy book "Functional analysis", page 145, has the formula $f(x)=\int_a^bK(x,y)f(y)\,dy=g(x)$, where $K$ is the kernel. This is fairly ubiquitous in functional analysis, in many variations.

My suspicion, just a hunch, is that the function analysis and algebraic cosets styles of kernels developed independently of each other historically, and then some algebra authors decided to apply the term to the equivalence relations which are induced by functions on their domains. In the earlier days, I'm sure that equivalence relations were presented without the word "kernel" to describe the equivalence relation induced by a function on its domain. But obviously, the algebraic definition of kernel is used to form quotient spaces which lead to isomorphism theorems in the same way that you get a quotient space from a general equivalence relation.

So my answer to the question is that the non-algebraic inverse-function style of kernel is just the obvious generalisation of the algebraic style. It isn't really "another" definition. It's just the natural extension to non-algebraic functions and quotient spaces.

But now I'm wondering what is the relation to the functional analysis style of kernel.

I've never seen this set referred to as a kernel, but I can give a positive answer to

Are there isomorphism theorems that use this language?

If $f\colon A\rightarrow B$ is just a plain set-function, then this kernel defines an equivalence on $A$, namely $$a\sim b\iff (a,b)\in \ker(f)\iff f(a)=f(b).$$ Then $f$ induces a function $\widetilde{f}$ on $A/{\sim}$ in the same way as other "first isomorphism theorems", for instance the one for groups, and $\widetilde{f}$ is a bijection (isomorphism) $A/{\sim}\rightarrow \operatorname{\mathrm{im}}(f)\subset B$.

The first isomorphism theorems for some algebraic structure follows from this by finding a way to give $A/{\sim}$ that algebraic structure and checking $\widetilde{f}$ respects this structure. In normal cases such as for, say, groups, $A/{\sim}$ is precisely $A/\ker(f)$, in more familiar sense of the kernel (and quotient). So I'd say you gain something in that using this definition allows you to do less work in the long run.

The typical "algebraic" definition of kernel only works for some structures. In particular, it does not make sense in the context of semi-groups, a perfectly fine algebraic structure where, due to the lack of an identity element, we cannot define the kernel of a homomorphism $h$ as $h^{-1}(\{e\})$.

Nor does it make sense in the context of topological spaces, where we have no multiplicative property to leverage, and yet we'd like to speak of quotient spaces in a way that mimics the algebraic construction.

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.