I'm having a little trouble understanding kernels and cokernels. Many authors refer to ''the'' kernel/coker of an arrow, when it's only unique up to an isomorphism on the domain for ker and codomain for cok. When we write ker(coker(f)), do we first mean a choice of coker and ker for every f? Sure we can write ker(f) and mean any kernel of f, but ker(coker(f)) presupposes that coker(f) has already been selected. And when we write m=ker(coker(m)) does it just mean we make a choice of ker/coker for each arrow, and m and ker(coker(m)) differ by an isomorphism on their domains that commutes with them (in the usual sense of subobject factorization)? Definitely in the category of abelian groups/modules we can make the usual canonical choice of ker/cok for each arrow, but a general abelian category doesn't have a "canonical" choice.
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$\begingroup$ see Mac Lane VIII.1 (or presumably any other standard source) $\endgroup$– user2035Commented Aug 8, 2010 at 8:51
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2$\begingroup$ A typical feature of the categorical properties is to define objects only up to isomorphism. THe various ker(f), coker(f) can't be distingished from any point of view, because every assertion made about a particular choice of ker(f) is valid, in the same form, for any other choice. $\endgroup$– foscoCommented Aug 8, 2010 at 9:02
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2$\begingroup$ You should remember, that the kernel (as well as the cokernel) is a morphism --- not just an object. Namely, a kernel of $f:X \to Y$ is a morphism $g:K \to X$ such that $f\circ g = 0$ and a universal property is satisfied. So, $Coker(Ker(f))$ is $Coker(g)$. $\endgroup$– SashaCommented Aug 8, 2010 at 10:37
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I guess not only different choices of kernel/cokernel are isomorphic but there is also a canonical isomorphism between any two. This is the case with any universal construction, e.g., direct product of objects. As far as I understand having defined an object up to a canonical isomorphism is as good as defining a canonical object.
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3$\begingroup$ Exactement. There is a natural commutative square (for any kernel & coker one chooses), using m along along top row, ker(coker(m)) along bottom, and specific vertical isoms. It is sloppy when people say something is unique "up to isomorphism", omitting canonicity and/or uniqueness of the isom (which underlies how the construction is actually used). In this sense, tetrapharmakon's comment is missing the canonicity aspect of the isomorphism, needed to avoid Amadeus' confusion. The end of Sasha's comment leads to Bill Clinton's contribution to math, on what the meaning of "is" is. :) $\endgroup$– BCnrdCommented Aug 8, 2010 at 15:38