$G/\mbox{Ker}\:\phi\cong H$ is a natural transformation?

Question

I was looking a bit into the subject of natural transformations, and I ended up in this discussion. The person giving this answer says that the transformation $G/\mbox{Ker}\:\phi\cong H$ is natural because it is a natural transformation between functors. What are the functors in this case?

Answer 1

One possible interpretation is that you look at the category $C$ whose objects are surjective group morphisms and whose arrows are the obvious commutative diagrams.

Then $F: (\phi : G\to H) \mapsto H$ is one functor, and $G: (\phi : G\to H)\mapsto G/\ker(\phi)$ is another one (check that you see how they are defined on arrows).

Then the canonical map $G/\ker(\phi)\to H$ is a natural transformation $F\to G$ evaluated at $(\phi : G\to H)$.

Note that surjectivity plays no role here, I just put it so that the transformation would actually be an isomorphism; but if you look at $\mathbf{Grp}^{\to}$, the category of arrows of $\mathbf{Grp}$, it works just as well, only you get a natural transformation $F\to G$ that is not an isomorphism.

Answer 2

Maybe this is one interpretation.

Let $\mathrm{Grp}$ be the category of groups with group homomorphisms, and let $\mathrm{Hom}_{\mathrm{Grp}}$ be the category of group homomorphisms $\pi:G\rightarrow H$. In this latter category, morphisms are defined as follows: if $\pi:G\rightarrow H$ and $\pi':G'\rightarrow H'$ are two group homomorphisms, then $\varphi:\pi\rightarrow \pi'$ consists of a pair of maps $\varphi=(g,h)$, where $g:G\rightarrow G'$ and $h:H\rightarrow H'$ are group homomorphisms so that the natural square commutes. Thus $\pi'\circ g=h\circ\pi$. (Sorry, I don't know how to tex diagrams on mathSE!)

Here are two functors $\mathrm{Hom}_\mathrm{Grp}\rightarrow \mathrm{Grp}$. One is the quotient by the kernel, the other is the image.

• Quotient-by-kernel functor: Define $\Phi: \mathrm{Hom}_\mathrm{Grp} \rightarrow \mathrm{Grp}$ as follows.

  • Objects: If $\pi:G\rightarrow H$ is a group homomorphism, set $\Phi(\pi):=G/\ker(\pi)$.
  • Morphisms: If $\varphi=(g,h)$ is a morphism from $\pi$ to $\pi'$, define $$\Phi:G/\ker(\pi)\rightarrow G'/\ker(\pi')$$ by $\Phi(\overline{x}):=\overline{g(x)}$, where $\overline{x}$ denotes passage to the appropriate quotient. This is actually well-defined because $\pi'g=h\pi$.

• Image functor: Define $\Psi:\mathrm{Hom}_\mathrm{Grp}\rightarrow\mathrm{Grp}$ as follows.

  • Objects: for $\pi:G\rightarrow H$, define $\Psi(\pi):=\mathrm{im}(\pi)$ to be the image of $\pi$.
  • Morphisms: If $\varphi=(g,h)$ is a morphism from $\pi$ to $\pi'$, define $$\Psi:\mathrm{im}(\pi)\rightarrow\mathrm{im}(\pi')$$ by $\Psi(\pi(x)):=\pi'(g(x))$. Once again you can check this is well-defined.

Now we have two functors $\Phi,\Psi:\mathrm{Hom}_\mathrm{Grp}\rightarrow\mathrm{Grp}$. Of course, these look the same, and essentially they are the same. Indeed we can define a natural transformation $$\eta:\Phi\rightarrow \Psi$$ by $\eta(\pi):G/\ker(\pi)\rightarrow \mathrm{im}(\pi)$ in the natural way.