2
$\begingroup$

Some of the work I have been doing lately is heavily dependent on chasing commutative diagrams so I have been brushing up on short exact sequences since I was not familiar with them. For the most part I feel I understand them but I have a couple things I want to clarify. In my examples the algebraic structures in my exact sequence's will be groups. In fact I think my misunderstandings might stem from the fact that I only have elementary knowledge of group theory.

Considering the short exact sequence

$$0 \rightarrow A \xrightarrow{\phi} B \xrightarrow{\psi} C \rightarrow 0$$

according to Wolfram

the group $A$ can be considered as a (normal) subgroup of $B$, and $C$ is isomorphic to $B/A$.

Most sites I checked made analogous statements. But what I in fact derived was only that $C \simeq B / \mathrm{Im(\phi)}$, but also that $\mathrm{Ker}(\psi) = \mathrm{Im(\phi)} \simeq A.$ So I feel I am very close to seeing what they are saying but something is holding me back. Since $A$ is isomorphic to Im($\Phi$) = Ker($\psi$) I see why they would say "$A$ can be thought of as a normal subgroup of $B$" since kernals are normal, specifically the normal subgroup such that $C \simeq B/A$. But I guess what I want to know is, what good does it do to $think$ about it being this way, even though it might not $actually$ be true that $C \simeq B/A$.

I have seen several examples where it is exactly true such as

$$0 \rightarrow 2\mathbb{Z} \hookrightarrow \mathbb{Z} \twoheadrightarrow \mathbb{Z} \, / \, 2\mathbb{Z} \rightarrow 0.$$

But for some reason that seems a bit contrived to me. Maybe because $C$ is exactly the quotient of $B$ and $A$ or because $A$ is exactly a normal subgroup of $B$.

I feel like one could easily find an example where it is not the case, although the general properties I mentioned I derived would still hold. What would this mean? Should you then just actually find the normal subgroup of $B$ that is isomorphic to $A$ and construct a new short exact sequence using this new group in place of $A$?

But there must really be something to this perception, even wiki says

It is helpful to think of A as a subobject of B giving rise to the isomorphism....

Why? Why is that helpful wiki?

Also, I add that I have already seen this similar question What is a short exact sequence telling me? but still felt my question had not been answered.

$\endgroup$
  • 1
    $\begingroup$ This way, some obtrusions calculated by homology can be stated as 'extensionality' problems: "How many homeomorphism from a submodule A, extend to the whole module B?", and you get answers like "if and only if the pushout of the A/B sequence along the homeomorphism is split". If one does not think in terms of submodules, one would state the theorem above as "the Ext of some modules A/B is zero iff...", which isn't necessarily bad, it's just less intuitive. $\endgroup$ – Andy Jul 16 '16 at 7:28
  • 3
    $\begingroup$ I think your problem is less about exact sequences than it is about replacing an object with another when they are isomorphic. When two groups $G_1$ and $G_2$ are isomorphic via a map $\phi$, every group theoritic property which is true for $G_1$ is also true for $G_2$. For example, if $x\in G_1$ is of order $n$, $\phi (x)$ too is of order $n$ in $G_2$. So you can really replace one with the other because $\phi$ is a perfect bridge. The same is true for rings, modules... $\endgroup$ – A.B. Jul 16 '16 at 7:36
  • $\begingroup$ Both comments appreciated. As I mentioned in a reply below I do think my understanding of this object is sufficient and my unease is something that will simply be cured by moving forward and getting more "used to" / experience in this area. Terminology question for @Andy - When you see "the pushout of the A/B sequence" .. I have seen on other sites people refer to a specific sequence in a short exact sequence. When you do this does it refer the map going into A/B? $\endgroup$ – Prince M Jul 16 '16 at 23:32
  • $\begingroup$ The pushout of a sequence A->B->C along a morphism f: A->D is another sequence: D-> PO -> C, where C is the same as before and PO is the pushout of A->B along f. If you have a map g: D -> C, you can instead pull back that map and have a sequence A -> PB -> D. Keep in mind that usually complexes are considered over an abelian category (such as modules over a ring) and the homology of groups is usually treated semi-separately because of the lack of 'abelianness' both in operations and categories. $\endgroup$ – Andy Jul 17 '16 at 7:42
3
$\begingroup$

I understand your confusion. Let me try and explain. Let $$ 0 \rightarrow A \xrightarrow{f} B \xrightarrow{g} C \rightarrow 0$$ be a short exact sequence of groups. Recall that exactness in this case is equivalent with saying $f$ is injective, the kernel of $g$ is the image of $f$ and $g$ is surjective. So $A \cong \text{Im}f \subset B$. So $A$ is isomorphic to a subgroup of $B$. So $B$ contains $A$ as a subgroup, up to isomorphism. In fact, we could replace every element of $A$ by its corresponding element of $\text{Im} f$, without changing anything structural. Intuitively I think of it like this: if two groups are isomorphic it means that they are the same 'up to symbols', that is, the one is another way of writing the other, but the group structure (the structure that is independent of how you write the group down) is the same. In this example, $B$ contains a subgroup that is 'essentially' $A$.

With the first isomorphism theorem we also have $B/\text{Im}f \cong C$. In fact, the short exact sequence above is equivalent to $$0 \rightarrow \text{Im} f \xrightarrow{\iota} B \xrightarrow{\varphi} B/\text{Im} \rightarrow 0$$ where $\iota$ is the injection map (it sends $a \in \text{Im} f$ to $a \in B$), and $\varphi$ is the canonical quotient map (it sends $b$ to $\overline{b}$). By the way, two short exact sequences are called equivalent if their corresponding groups are isomorphic and the diagram induced by these isomorphisms commutes. I hope this makes things a bit more clear.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I appreciate your answer, especially since I didn't really purpose a specific question, but this is what I was looking for. It is reassuring that my take on the S.E.S is consistent with your answer which tells me my understanding is sufficient for now and I should move forward, as they arise naturally in my work it will probably fill in the small gaps of uncertainty that remain. That is how it has been for most objects that I was uncertain about at first! $\endgroup$ – Prince M Jul 16 '16 at 23:25
  • $\begingroup$ For now, do you have an elementary example of how a short exact sequence might arise, be used to solve a problem, be used in a proof.. etc? Perhaps still in the language of groups for simplicity? @M. Van $\endgroup$ – Prince M Jul 16 '16 at 23:27
  • 1
    $\begingroup$ I find that thinking of the inverse problem (which is particularly rich in the case of groups) is useful: you start with two groups A and C; how many groups B are there such that C = B/A? It is equivalent to know how many A->B->C exact sequences there are. Surely the direct sum will always work, all semidirect products would work...etc. $\endgroup$ – Andy Jul 17 '16 at 7:50
  • 1
    $\begingroup$ I wanted to say exactly what Andy said :) $\endgroup$ – M. Van Jul 17 '16 at 13:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.