# Category theory: Prove that two given short Ex. Seq.s are isomorphic.

1. Mar 30, 2012

I've attached the problem statement and the solution in a pdf file, please check it and see if I've done it correctly. I'm new to abstract reasoning, I've only had a one semester introduction to group-theory and parts of ring theory based on baby Herstein, so I need others to check my proofs to know whether I'm doing them right or wrong.

#### Attached Files:

• ###### Example 9.pdf
File size:
833.2 KB
Views:
122
2. Mar 30, 2012

### jgens

A couple of points:
1. You need to prove that the diagram to commutes to prove that you have an isomorphism of short exact sequences. This is currently the biggest problem with your argument.
2. Notice that $\psi$ is only required to be surjective, so writing something like $\gamma = \pi\psi^{-1}$ is nonsense. You explain later what you mean, and you have the right idea, so this is just a notational issue.
3. Proving that $\alpha,\beta,\gamma$ are isomorphisms is much easier than you think. Defining $\alpha:M \rightarrow \mathrm{im}\,\phi$ such that $\alpha(x)=\phi(x)$ forces $\alpha$ to be injective since $\phi$ is injective by hypothesis and proving surjectivity is similarly easy. Defining $\beta = \mathrm{id}_N$ is clearly an isomorphism. Lastly there exists a canonical isomorphism $\gamma:K \rightarrow N/\ker \psi$ by the first isomorphism theorem for modules. This is the same map you define and it saves you the trouble of proving that $\gamma$ is well-defined.
Try rewriting it with these concerns in mind.

3. Mar 31, 2012

Well, We create the maps in a way that the diagram commutes, that's how I obtain α
,β and γ. You're right though, because I haven't precisely shown that γψ=π (It's really very obvious that the rest of the diagram commute) but it's not immediately followed from γ
=πψ-1 that we have γψ=π, so I'll fix this hole in my argument later.

Yea, you're right, using the notation ψ-1 might cause the confusion that ψ
is bijective, but ψ-1 isn't really the inverse of ψ, it's only the right inverse which exists because ψ is surjective, so I should've used another letter for the right inverse of ψ, like another Greek letter or whatever.

That is exactly what I've done.
Yup, I could use the first isomorphism theorem as well because $\ker \psi$=$Im \varphi$, that would've reduced the exhausting work to only one line.

4. Mar 31, 2012

### jgens

1. The point is to construct the relevant isomorphisms and prove that they commute; the fact that it is "obvious" in this case is irrelevant since the argument itself is pretty obvious. You also need to define your isomorphisms first before you prove that anything commutes. Doing it any other way is poor form. It is generally a good idea to get some practice with diagram chasing anyway.

2. Do not use right inverses to construct your isomorphism $/gamma$. There is a canonical map which exists by the first isomorphism theorem and this works fine. So there is no need to even mention right inverses anywhere in your proof.

3. Streamline your proof. Unless you are writing this proof for pedagogical reasons, you are much too wordy. There is no need to explain why you chose each isomorphism or anything like that. Just define them in one paragraph and prove everything commutes in the next paragraph. Learning how to write clean proofs is a useful skill.