Definition of kernel Is the kernel the same as the union of all level sets (or perhaps the set of all level sets)?
Kernel (https://en.m.wikipedia.org/wiki/Kernel_(set_theory)):
$\left\{\, \left\{\, w \in X \mid f(x)=f(w) \,\right\}  \mid x \in X \,\right\}$ 
Level set (https://en.m.wikipedia.org/wiki/Level_set)
$L_c(f) = \left\{ (x_1, \cdots, x_n) \, \mid \, f(x_1, \cdots, x_n) = c \right\}~,$
Why is the kernel described as the vectors that are mapped to zero in vector spaces? Surely there are more level sets than that?
 A: The kernel of a linear map $f$ is a level set, yes. And it is a particulary important one, since $\ker f=\{0\}\iff f$ is injective.
A: The kernel for general maps between sets is an equivalence relation: if $f\colon X\to Y$, then the kernel is the equivalence relation $\sim_f$ defined by
$$
a\sim_f b\text{ if and only if }f(a)=f(b)
$$
The Wikipedia page identifies this relation with the partition induced by it:
$$
\bigl\{ \{w\in X:f(x)=f(w)\}:x\in X \bigr\}
$$
where, for $x\in X$, $\{w\in X:f(x)=f(w)\}=\{w\in X:x\sim_f w\}$ is the equivalence class (or level set) of $x$.
When linear maps are concerned, there's a better description, because when $X$ and $Y$ are vector spaces and $f$ is linear,
$$
f(a)=f(b)\text{ if and only if }f(a-b)=0
$$
so the kernel can be described just by the vector subspace $N=\{x\in X:f(x)=0\}$.
There's no real difference, except that in vector spaces (but also in groups or rings) the description of the kernel is handier.
The key fact is that $\sim_f$ is more than an equivalence relation: it is a congruence, that is, an equivalence relation that preserves the operations on the structure: if $a\sim_f b$ and $a'\sim_f b'$, then
$$
a+a'\sim_f b+b'
$$
and, when $\gamma$ is a scalar, also $\gamma a\sim_f \gamma b$.
The fact that a vector space (and likewise a group or a ring) has an operation which is associative, with neutral element and inverses, allows for the simpler description in terms of a single subset rather than with a partition.
