jgens said:
[*]You need to prove that the diagram commutes to prove that you have an isomorphism of short exact sequences. This is currently the biggest problem with your argument.
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.
[*]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.
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.
[*]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.
That is exactly what I've done.
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.
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.