Among different definitions of kernels and cokernels I wonder if anybody tried to study the following construction.
Let $\mathcal K$ be an arbitrary category (there is no need to require the existence of a zero, or of a system of zero morphisms).
- Let us say that a morphism $\psi:K\to B$ is trivial from the inside (or constant, or left zero), if for any two parallel morphisms $\eta,\theta:X\to K$ we have $$ \psi\circ\eta=\psi\circ\theta. $$
- Let us say that a morphism $\varkappa:K\to A$ trivializes a morphism $\varphi:A\to B$ from the inside, if their composition $\varphi\circ\varkappa:K\to B$ is trivial from the inside.
- Let us say that a morphism $\varkappa:K\to A$ is a kernel of a morphism $\varphi:A\to B$, if $\varkappa$ trivializes $\varphi$ from the inside, and for any other morphism $\varkappa':K'\to A$ that trivializes $\varphi$ from the inside, there is a unique morphism $\iota:K'\to K$ such that $\varkappa\circ\iota=\varkappa'$.
Dually, we define the notion of cokernel.
It is easy to see that if the category $\mathcal K$ has zero, then these definitions give the usual kernel and cokernel.
However, I can't find categories with kernels and cokernels in this sense, were zero does not exist, all examples that come to my mind are categories with zero. So my question is the following:
Are there categories without zero, where kernels and cokernels in this sense exist for each morphism $\varphi:A\to B$?
I would appreciate if somebody could give examples of popular categories, that are used in mathematics, but even methodical examples are interesting to me.
P.S. I am checking which categories have kernels and cokernels in this sense. As I see, the category $_k{\sf Alg}$ of algebras over a field $k$ has no kernels, but I faced difficulties in proving the same about cokernels. Perhaps, somebody knows an example?
P.P.S. From Martin Brandenburg's answer below it follows that the category $_k{\sf Alg}$ of algebras over a field $k$ has no cokernels. As an example, we can take the morphisms $\eta,\theta:k^2\to k$, $\eta(s,t)=s$, $\theta(s,t)=t$, they are trivial from the outside, but their "product" $\eta\sqcap\theta=1_{k^2}:k^2\to k^2=k\sqcap k$ is not trivial from the outside.
Here is why often this sort of kernel does not exist.
By your definition, the kernel of a morphism $ A \to B$ is a representation of the subfunctor of $\hom(-,A)$ that consists of those morphisms $K \to A$ such that the composition $K \to A \to B$ is constant*. (This is completely analogous to the classical definition of a kernel when zero morphisms are present; then one demands that $K \to A \to B$ is a zero morphism.) For simplicity, let us look at the identity morphism $A \to A$. Then we get the subfunctor of $\hom(-,A)$ that consists of constant morphisms, and $A \to A$ has a kernel iff this subfunctor is representable. Let's call it $F$.
Now, every representable functor preserves all limits (and up to set-theoretical issues, and completeness of the source category, this is all we need according to Freyd's criterion for representability). Here, since we have a contravariant functor $F : \mathcal{K}^{\mathrm{op}} \to \mathbf{Set}$, this implies that for every family of objects $(X_i)_{i \in I}$ in $\mathcal{K}$ that have a coproduct $\coprod_{i \in I} X_i$, the canonical map $$\textstyle F\bigl(\coprod_{i \in I} X_i\bigr) \to \prod_{i \in I} F(X_i)$$ is an isomorphism, i.e. bijective. It is injective anyway, and surjectivity says that for every family of constant morphisms $X_i \to A$ the induced morphism $\coprod_{i \in I} X_i \to A$ is also constant. This is obviously wrong for the category of sets, for example.
Let us consider the special case of $X \sqcup X$, the coproduct of two copies of $X$. Take two constant morphisms $f,g : X \to A$. Then $(f,g) : X \sqcup X \to A$ is supposed to be constant. In particular, $(f,g) \circ \iota_1 = (f,g) \circ \iota_2$, which means $f = g$.
So, assuming the existence of binary coproducts and the existence of kernels, every two parallel constant morphisms will be equal. This is almost already a family of zero morphisms (which are also unique), except that here maybe there is no constant morphism at all.
*The conceptually correct definition of a constant map of sets $f : A \to B$ is that there is some $b \in B$ such that $f(a) = b$ for all $a \in A$. More abstractly, $f$ is constant when it factors through the terminal object. This implies that the unique map of sets $\emptyset \to \emptyset$ is not constant. Your definition, though, would call this map constant. A map satisfying $f(a)=f(a')$ for all $a,a' \in A$ could be called "weakly constant" or something like that. But I have followed your terminology above and just call these maps constant.