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I'm reading Mac Lane's "Categories for the Working Mathematician". I found the following sentence in page 59 of it:

Similarly, the kernel of a homomorphism (in $\mathbf{Ab}$, $\mathbf{Grp}$, $\mathbf{Rng}$, $R$-$\mathbf{Mod}$, . . .) is a universal, more exactly, a universal for a suitable contravariant functor.

I know how to express a quotient by a universal property in a category. But I really have trouble understanding how to express the kernel of a homomorphism as having a universal property.

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  • $\begingroup$ Try this: Reverse all arrows in what you have done for the quotient. $\endgroup$ – c_c_chaos Apr 9 '15 at 13:07
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Let $f : G \to H$ be a homomorphism of abelian groups. The kernel of $f$ is a pair $(K,i)$, consisting of an abelian group $K$, namely $\{g \in G : f(g)=0\}$, and a homomorphism $i : K \to G$, the inclusion, with $fi = 0$. Here, I denote with $0$ the trivial homomorphism. If $(T,j)$ is another such pair, there is a unique homomorphism $k : T \to K$ such that $ik=j$. (Draw a diagram!) This is because $f(j(t))=0$ implies $j(t) \in K$, etc. This is the universal property of the kernel.

How to interpret this as a universal element of a functor? Well, consider the functor $\mathsf{Grp}^{\mathrm{op}} \to \mathsf{Set}$ which maps to each group $T$ the set of all homomorphisms $j : T \to G$ with $fj=0$. The action on morphisms is clear. In fact, this will be a subfunctor of $\hom(-,G)$. Then, the kernel $(K,i)$ is a universal element of this functor.

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  • $\begingroup$ Thank you very much, now just a little problem remained. In Mac Lane's book, $\mathbf{Rng}$ denotes the categories of unital rings. So there is no trivial map in $\mathbf{Rng}$ I think, and an ideal need not lie in $\mathrm{ob}\ \mathbf{Rng}$. Could you please tell me how to make your method applicable for $\mathbf{Rng}$? $\endgroup$ – Censi LI Apr 9 '15 at 14:07
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    $\begingroup$ The category of rings has no kernels (in any meaningful sense). But the category of non-unital rings has kernels. I don't know what S. Mac Lane intended. $\endgroup$ – Martin Brandenburg Apr 9 '15 at 14:14
  • $\begingroup$ It's not possible to give an ideal any categorical meaning to be a kernel of a morphism in $\mathbf{Rng}$? $\endgroup$ – Censi LI Apr 9 '15 at 14:36
  • $\begingroup$ Yes. There are some ways to do this, using the forgetful functor to $\mathsf{Ab}$, or the slice category $\mathsf{Rng}/\mathbb{Z}$, which is equivalent to the category of non-unital rings, but I don't think that you can stay within $\mathsf{Rng}$. $\endgroup$ – Martin Brandenburg Apr 9 '15 at 15:03
  • $\begingroup$ A good substitute is the kernel pair of a ring homomorphism $f : R \to S$, which is the subring of $R \times R$ consisting of those elements $(x,y)$ with $f(x)=f(y)$. For example, $f$ is injective if and only if the kernel pair of $f$ is trivial in the sense that it equals the diagonal subring $\Delta_R$. Of course, this has nothing to do with rings - it works for all algebraic structures. $\endgroup$ – Martin Brandenburg Apr 9 '15 at 15:05
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In an abelian category, a kernel of a morphism $f\colon X\to Y$ is a solution of the following universal problem: find a morphism $i\colon K\to X$ such that:

  1. $f\mkern 1.5mu i=0$.
  2. for any morphism $u\colon Z\to X$ such that $f\mkern 1.5mu u=0$, there exists a morphism $v:Z\to K$ such that $u=i\mkern 1.5mu v$.

More generally, in a category with zero morphisms, $\ker f$ can be defined as the equaliser of $f$ and $\, 0_{XY}$ (again a universal problem).

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  • $\begingroup$ Thank you very much, now just a little problem remained. In Mac Lane's book, $\mathbf{Rng}$ denotes the categories of unital rings. So there is no zero morphisms in $\mathbf{Rng}$ I think. And $\ker f$ need not lie in $\mathrm{ob}\ \mathbf{Rng}$. Could you please tell me how to make your method applicable for $\mathbf{Rng}$? $\endgroup$ – Censi LI Apr 9 '15 at 14:09
  • $\begingroup$ Not sure, but I suppose you have to embed Rng in Ab, or maybe consider the category X-Mod (X being the source of $f$). $\endgroup$ – Bernard Apr 9 '15 at 14:37

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