Define a monomorphism (epimorphism) as a left-cancellative (right-cancellative) morphism, in some category.
The Wikipedia page on cokernels says that a morphism is injective (surjective) iff its kernel (cokernel) is trivial. The category being discussed seems to be that of vector spaces.
Now, in many categories, being an injection (surjection) is equivalent to being a monomorphism (epimorphism), but, if this is not the case –ie. monomorphisms are not injective (epimorphisms are not surjective)–, does the above result generalize to such monomorphisms (epimorphisms)?
Namely, is the following true? A morphism is a monomorphism (epimorphism) iff its kernel (cokernel) is trivial.
Best Answer
As Qiaochu notes, kernels and cokernels don't make sense in an arbitrary category: you need a zero object to even define what these should be (and even with a zero object, they aren't guaranteed to exist).
If you have a category with a zero object, then a monomorphism $f$ has zero kernel: $f\circ\ker f = 0 = f\circ 0\implies \ker f = 0$. This doesn't show that the kernel actually exists, just that if it does, it must be equal to zero. However, it is not difficult to verify that $0$ satisfies the universal property of the kernel in this case. Reverse all the arrows to get the dual statement for cokernels.
However, the converse is false: consider the category with four objects $A$, $B$, $C$, and a zero object $0$ and morphisms \begin{align*} \operatorname{Hom}(X,X) &= \{0, id_X\}\quad\textrm{for each object }X\\ \operatorname{Hom}(B,C) &= \{0, f\}\\ \operatorname{Hom}(A,B) &= \{0,g,h\}\\ \operatorname{Hom}(A,C) &= \{0, fg = fh\}\\ \operatorname{Hom}(C,A) &= \operatorname{Hom}(C,B) = \operatorname{Hom}(B,A) =\{0\}. \end{align*} By construction, $g\neq h$, $fg = fh$, and you can verify that $\ker f$ exists and is equal to $0 : 0\to X$.