Relationship between congruence and kernel of homomorphism

Question

First the terms as I understand them:

1. The congruence is defined as equivalence that preserves structure (the operations) of algebra.

2. Homomorphism is a map f: A->B between two algebraic structures of the same type, that preserves structure. This means that it doesn't matter if we map the result of operation, or map the operands and then perform operation. For example f(a+b) = f(a)+f(b); a,b \in A.

3. By this definition: Kernel of homomorphism is a set of elements from A which are mapped to neutral element of B. For example if the map is from (R^2,*) -> (R,*) (2D vectors to real number) defined as f(x) = |x| (length of vector), all vectors that would map to 1 (which is neutral in (R,*)) would make up the kernel. Therefore the kernel is set of all unit vectors.

4. From other definition: Generally, the kernel is a congruence relation.

Regarding this, I have following questions:

1. Is there anything wrong with the definitions or examples above?
2. How can it be, that generally the kernel of map is a congruence, but kernel of homomorphism is a set of elements from A? Shouldn't be these 2 distinct? What is the relationship between them?
3. How would a quotient algebra defined as A/(kerF), where kerF is kernel of homomorphism look like in example from third definition? How is it different from quotient algebra where kerF is general kernel from fourth definition?

Answer 1

In "nonlinear" contexts the correct replacement of the kernel is a construction called the kernel pair. The kernel pair of a morphism $f : X \to Y$ is abstractly the pullback $X \times_Y X$. Concretely, in familiar concrete categories, it can be described as the subset

$$\{ (x, x') \in X \times X : f(x) = f(x') \}$$

of $X$. This subset describes an equivalence relation on $X$, and in the case of a morphism $f : G \to H$ of groups, the equivalence relation is that $x$ and $x'$ differ by an element of the kernel $\text{ker}(f)$ in the usual sense. With a little more effort you can show that this sets up a natural bijection between congruences on a group and normal subgroups; see this blog post for some details.

The kernel pair is equipped with two projection morphisms to $X$, and the correct replacement of quotienting by the kernel is taking their coequalizer.