let's see who on mastodon is in the 2%!!!!

@hierarchon I fucking give up, the answer is Deez nuts anyway

@hierarchon uhh hold on let me re-read my algebra texts and i'll get back to you

@KitRedgrave @hierarchon I suspect you need some post-graduate mathematics to even understand the question being asked here. I have a minor in math and I'm sitting here going "I recognize most of these words and symbols but have no idea why they're in that order"

@hierarchon @KitRedgrave *pops knuckles delicately* Arright.

@hierarchon @KitRedgrave Firstably how dare you have a hamburger and not use it for the Hom set

@CoronaCoreanici @KitRedgrave i didn't make this, i just found it on some image search while i was looking for something else :v

@hierarchon @KitRedgrave OK, got nerd-sniped to fuck, and here's my conclusion

whoever wrote this is an asshole, because the index variable i in the exact sequence isn't actually specified anywhere, and I don't think this actually holds for all possible i (try i = n+1, for instance - the middle term vanishes, but the second and fourth don't, so this can't be exact)

@hierarchon @KitRedgrave Admittedly I'm not 100% sure that $\banana$ is actually the Tor functor. Also, $\mathscr{C}$ isn't actually defined anywhere, either, so unless that's meant to be a comment that Hom-ing into Set from any other category is contravariant, this problem has some issues

@hierarchon @KitRedgrave ...hush, abstract predators are allowed to show their power level little a, as a treat.

@CoronaCoreanici @KitRedgrave i don't know nearly enough about this kind of thing as i want to, but i never get around to learning

@CoronaCoreanici @hierarchon @KitRedgrave yeah that's what I'm confused about, a lot of stuff isn't defined at all. I'm not convinced it's even asking a well-formed question(although I'm also garbage at anything to do with category theory)

@zardoz @hierarchon @KitRedgrave The question it's asking is actually decently well-formed apart from that. It's asking about the nature of a graded ring over Z/2 in terms of the nth cohomology groups of (I think?) the real projective n-space. (the dimension of which is ALSO unspecified)

@CoronaCoreanici @hierarchon @KitRedgrave yeah but there's also all that shit about H that looks completely pointless

@CoronaCoreanici @zardoz @hierarchon @KitRedgrave hmm, yes, from this i can deduce that these are definitely words referring to concepts, indeed

@CoronaCoreanici @hierarchon @KitRedgrave I think most of it is superfluous information though. The object they're asking about isn't very complicated.

@zardoz @hierarchon @KitRedgrave Yeah, now that I think about it, the exact sequence isn't used anywhere in the actual question, and neither is Tor.

@CoronaCoreanici @hierarchon @KitRedgrave I'm gonna take a total shot in the dark and say that the answer is (Z/2Z)[x]/<x^n+1>

@zardoz @hierarchon @KitRedgrave That definitely can't be right - they're looking for a graded cohomology ring, and that'd have to have indexed components over k \in \nat.

@CoronaCoreanici @hierarchon @KitRedgrave isn't any polynomial ring graded?

@zardoz @hierarchon @KitRedgrave Oh, wait, are you saying that the ring is \oplus_{n \in \nat} (Z/2)[x]/<x^n + 1>?

@CoronaCoreanici @hierarchon @KitRedgrave no, just (Z/2)[x]/<x^n + 1>

plus a bunch of zero rings because H^N (S,R) = 0 for N > dim(S)

@CoronaCoreanici @hierarchon @KitRedgrave (fwiw I looked up the answer and this is definitely wrong)

@zardoz @hierarchon @KitRedgrave Ah, I see. But won't some of the homologies for RP^n in Z/2 coefficients be nontrivial for k < n?

@CoronaCoreanici @hierarchon @KitRedgrave yeah. The individual cohomology groups would be {0,x^i} for some fixed i

@zardoz @hierarchon @KitRedgrave Then the direct sum of all of those would be your graded ring of cohomology rings... I think? There's something to check about the actual grading on the ring but that looks to follow immediately.

@CoronaCoreanici @zardoz @hierarchon @KitRedgrave
annoyingly I think the sequence given is stated incorrecly. It's the universal coefficient sequence for cohomology but the role of the burger and the sandwich should be swapped

@CoronaCoreanici @zardoz @hierarchon @KitRedgrave all I know is that Thor has a hammer.

@hierarchon Hah, such an elementary problem is nothing to my advanced intellect.

However, it would be unfair of me to spoil the answer. I will leave it to the rest of you to sort it out.

@Anarkat
No, no, my friend. You need to use the correct terminology.

"The solution is left as an exercise to the reader."
@hierarchon

@hierarchon uhhh 69420?

@hierarchon *brought to tears by the beauty of mathematics*

@hierarchon oh, i love* the putnam

* was afraid to even try taking

@hierarchon

@hierarchon i can feel my brain seeping out of my nostrils

@hierarchon i know for a fact that burger is 1/2 but that's as far as I got

@hierarchon contravariant? derived functor? When did this leave math and enter programming?

@moonbolt the way mathematicians use 'functor' is very different from the way the c++ nerds use it (but closer to how the haskell nerds use it)

@hierarchon wait, C++ has functors?

@moonbolt a 'functor' in c++ means an object you can call (i.e., that implements operator())

@hierarchon that's just /function/ in the broad sense, or /callable/.

@moonbolt i never said it was a good name!

@hierarchon what is this, send help

@hierarchon this hurts

H^*(🌭;🍔) = 🍔[🍟]/(🍟^(n+1))

where 🍟 is of degree 🍇

@hierarchon
I gave up after hotdog = P^n(R).

@hierarchon delet this >:(

@hierarchon Okay listen I'm dumb as fuck so while I can appreciate the surface-level joke, is there another, secret layer where this is actually a solvable thing? Or is it just nonsense?

@hierarchon it has no solution because cookie is never defined and could be zero, which would cause hamburger to also be undefined.

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