As I read in this problem Confusion about categorical viewpoint of normal subgroups, it's said that
Let me state the two definitions a bit more explicitly. Let $\mathcal{C}$ be a category with a zero object. For any objects $A$ and $B$, we let $0:A\to B$ denote the unique map which factors through a zero object. Then:
- A normal monomorphism is a map $i:A\to B$ such that there exists an object $C$ and a map $f:B\to C$ such that $i$ is the equalizer of $f$ and $0:B\to C$.
I find this defination sounds reasonable, however, it is difficult to me to apply it. For example, if I want to show that $U$ and $W$ is normal subgroup of $U\times W$, it's not easy, since I cannot see any reason to combine the universal property of product and the universal property of equalizer. Can you show me how to prove this property ?
Meanwhile, I find normal subgroup has more deeper means in terms of split short exact sequence since if both $H$ and $N$ are normal subgroup, the spliting lemma will work, that's to say $1\to N\to G\to H\to 1$ is split, then $G = N\times H$. So can this be explained by the universal property above ?
You want to find a normal monomorphism $U\to U\times W$ and find a suitable object $C$. Let’s be concrete for the moment; firstly, due to the existence of a quotient construction, normal subgroups are exactly the same thing as kernels which are exactly the same thing as normal monomorphisms (easy exercise: kernels are equalisers of $0$ and the relevant homomorphism). Secondly, ‘the’ way to view $U$ as a subgroup of $U\times W=U\oplus W$ is by identifying it with $\{(u,0):u\in U\}$. Thinking categorically, that means we care about the monomorphism $U\to U\times W$ which takes $u$ to $(u,0)$. Using the universal property of a product we can encode this purely categorically as: $$i:=\langle\mathrm{id}_U,0_{WV}\rangle:U\to U\times W$$
This is a monomorphism because if $if=ig$ then $\pi_1 if=\pi_1 ig$, that is $\mathrm{id}_U f=\mathrm{id}_U g$ do $f=g$ (note we kept that proof categorical, without appeal to elements of the group). Is it a normal monomorphism? Well, thinking concretely again we should remember that the obvious map which makes $U\times0$ a kernel is the second projection $\pi_2:U\times W\to W$; I’m sure every student is asked at some point to check $U\times W/U\cong W$. So, take $C=W$ and $f=\pi_2$.
Firstly, $\pi_2 i=0$ by definition (using universal properties!) of the map $i$. If $\pi_2 h=0$ for $h:X\to U\times W$, the universal product property allows us to uniquely encode $h$ as $\langle h_1,h_2\rangle$ and by definition the second component is zero; $h_2=0$. Using universal properties again, it’s not too hard to check $ih_1=h$; since $i$ is a monomorphism, that makes $h_1$ the unique arrow $X\to U$ as per the definition of equaliser. Done!
Notice that was unnecessarily laborious. When dealing with groups I think it’s safe to say that using the concrete properties and explicit constructions usually makes your life easier without much loss of insight, certainly so for checking if subgroups are normal.
As for the splitting lemma, it’s easy with Abelian category theory (note; the category of groups is not Abelian… but I mention this only to say that these things can be checked without appeal to elements in categorical fashion) to check that, since $N\to G$ is forced to be monic, it is isomorphic to its image object and the exactness condition implies $N$ itself is ‘the’ (a) kernel of $G\to H$; this is the same thing as saying $N\to G$ is a normal monomorphism, taking $C=H$. So regardless of whether or not that sequence splits, $N$ will have to be identifiable with a normal subgroup of $G$. It’s then just the abstract splitting lemma that $G\cong N\oplus H\cong N\times H$, using the fact that the map $G\to H$ admits an right inverse and the map $N\to G$ admits a left inverse (which we take as the product projection $G\to N$).