I am trying to understand what exactly 'ker f' is. My guess is that it is a set of all of the elements that were lost in the process of a mapping.
For example: $f:A\to B$
$f = \left\{ \begin{align} \frac x2, \text{ if } x = 0 \pmod 2 \\0, \text{ if } x = 1 \pmod 2 \end{align} \right.$
Here the kernel of $f$ would contain all the values that were odd before the mapping. Is this correct?
$\endgroup$ 32 Answers
$\begingroup$Before I start, I should say that the function you've given, if it is from $\mathbb Z\to\mathbb Z$, is not a homomorphism: $$f(3)+f(5)=0+0=0 ~~\neq~~ 4 = f(3+5).$$
I am not used to seeing $\ker(f)$ being used for functions other than homomorphisms, but it is possible (if a little silly) to extend the usual definition for arbitrary functions $f:X\to G$, where $G$ is a group.
The usual definition for the kernel of a group homomorphism $f:G\to H$ is that $\ker(f)$ consists of all those elements in the domain which are mapped to the identity of the codomain. That is, $\{x\in G:f(x)=e_H\}$. You can of course imagine this being meaningful even when the domain is not a group, but it does not make sense when the codomain is not a group.
However, in such a case we are generally not concerned with kernels, because it turns out that what we really want a kernel to do is be a special kind of subgroup of the domain. But of course if the domain is not a group, then this notion doesn't even make sense.
[ Here is a useful exercise: prove that the above application is meaningful. In other words, show that when the domain is a group, then $\ker(f)$ really is a subgroup of its domain. And a brief followup: what does it mean if $\ker(f)$ turns out to be the trivial group? ]
[ Actually, there is an extended definition of "kernel" that would cover generic functions $X\to Y$, but I don't want to introduce any confusion. ]
$\endgroup$ 2 $\begingroup$Not quite. The definition is that $\text{ker} f = \{x \in A | f(x) = 0\}$.
Since $0 \equiv 0\pmod{2}$, we have $f(0) = \dfrac{0}{2} = 0$, so $0 \in \text{ker} f$.
The kernel of $f$ consists of all odd integers and $0$.
$\endgroup$
