Daniel Donnelly

I have proported twin prime conjecture proof. Attempt number 99

Btw arrows = homomorphisms here

So abstraction is good but not at the expense of not being able to actually do the math. Which is why most diagram chase proofs let you take elements at the nodes (abelian groups)

So you have the category $\textbf{Ab}$ of abelian groups. This forms an abelian category. In a general abelian category there is no notion of elements other than what you can model with arrows ("generalized elements"). However because of Freyd-Mitchell embedding theorem you may do diagram chases in some category of $R$-modules (essentially vector spaces without invertibility of scalars) and that suffices to do it in your original abelian category (very general / abstract)

They call that pure arrow-based math

Some things can be proven without examining the elements but the proofs seem very alien to me and hard to find

In category theory they've abstracted this a little

Homomorphism just means structure preserving map in algebra

No worries 😵

Not sure what you mean by that, pls explain

They sometimes use it in some proofs as a way to do induction because $G/N$ is usually strictly smaller in cardinality than $G$

Not sure, quotient groups are broadly applicable to every area of math

It's like how modular arithmetic works because that's a special case of this general phenomenon

You can work equivalently in $G$ or in $G/N$ in order to work in $G/N$ but after each computation in $G$ you need to "modulo by $N$" in order to get your result in $G/N$.

I think I know what your question was about though.

No that's the second thing. Once you have well-definedness proven, you have to prove that your function is actually a group hom

It's used all the time though, say in HA (homological algebra)

Well-definedness simply means that it's actually a function and agrees on any choice of representative

It's just namely well-definedness out of the quotient that can fail

All that needs to be proven by you though before you become confident

I mean $h \circ f$

So essentially, the mnemonic is this: Given a group (ring, module, etc) homomorphism $g : G \to H$, if you have a (normal) subgroup of the kernel of $g$, then you may factor $g$ through it! Forming $g = f \circ h$, where $f: G \to G/K$ and $h : G/K \to H$.

Project I'm working on

ANY TIME you have a quotient taken, you always have to prove that maps FROM IT are well-defined.

Use that to prove that if $h + K = g + K$ (two reps of the same equivalence class) then $f(h) = f(g)$ where $f : G/K \to G/N$ is the map we're trying to prove well-defined.

This is true since $\pi$ takes the $K$ to zero (being a subset of the kernel $N$ of $\pi$)

This is because of well-definedness of $g + K \mapsto \pi(g)$ where $\pi : G \to G/N$ is your quotient natural projection

I mean $G/K \to G/N$ is well defined whenever $K \leqslant N \leqslant G$ all normal

I'm confusing myself though, so take this with a grain of salt

@X4J Any time you have a normal subgroup of your normal subgroup, you may factor through the first one as the natural map you define becomes well-defined. That's used a lot in advanced topics.

Evening

But I can't crack it

@ModularMindset I upvoted

What could you do with reversible homology measurements?

@ModularMindset what is that algebraic topology?

@GerryMyerson, I have a NT attack on several problems. Approach is elementary though, so who's gonna read it, as you seem to imply. The results get into deep areas such as Group Theory, but don't use that much of it.

I think to answer this question it would be helpful for us each to pick the well-known open problem we've secretly or not so secretly have been working on in our spare time, and show with first-hand knowledge why the problem is so difficult. Like this tried list of attacks doesn't work, and here's the wall you run into. Hence my edits.

@GerryMyerson define important in this context.

Anyone up to chat homology?

It's the responsibility of mathematicians as a whole to explore what the implications are. You can't expect one person to have the whole new theory laid out - it's too much work, especially to go into the question of a Q&A site. Therefore, I ask if any farther afield experts in the area of HA have something to say about it.

I'm up for more attacks against the statement. All of sudden generalization is a bad thing.

@JohnPalmieri you take each HA proposition, and see where you can get away with reverse homology instead of full-fledged exactness. You can then operate on reverse and forward sequences as units, but you can't mix forward and reverse in the same sequence and retain an abelian category because the kernel/image containment direction can be reversed randomly, so is not fixed to one and so there is no consistent direction. I'm not sure if there could be a morphism between forward and reverse complexes ("sequences")

@JohnPalmieri added Down2Earth example at the bottom.

The theory generalizes in more places than one and the Snake Lemma is used everywhere like with Yoneda's Lemma in category theory. Let its effect on generalizing the entire theory speak for itself.

@BenSteffan a tool that can do much more now.

@NoahSchweber when do mathematicians not care about generalizing a theorem?

@JohnPalmieri essentially anywhere you rely on exactness (as a containment) in a proof but not on $d^2 = 0$ equation itself, then the theorem can be generalized in some way to include sequences with reverse homology.

In the SRS to LES it should be LRS as the final sequence could have reversed homology in some of the terms.

In fact you just say "having reversed (row-wise) homology" in each node of the usual snake lemma, and you have a generalization of it that covers all the above mentioned cases.