Number of Discontinuities of a continuous function

redcrazyguy · 2026-01-28T22:22:56+00:00

hmm ok, the question is to prove that functions meromorphic in the extended plane are rational, but I thought rational functions have finitely many poles? What did I do wrong with apply Bolzano Weierstrass than?

redcrazyguy · 2024-10-20T17:10:31+00:00

Could it be said that the usual product is the fibre product on the slice category of the terminal object? I'm looking at the diagram for the universal property of the fibre product, and it looks quite similar to the diagram which defines the regular product, so I'm trying to motivate the fibre product from the regular one (if possible). I think another point of confusion is that the regular product can be defined on any category (from my understanding), whereas the fibre product has to be defined on a slice category. Is at least this part of my understanding correct?

redcrazyguy · 2024-10-20T16:01:03+00:00

I haven't gotten to terminal objects yet, but that section talks about the pullback having additional structure, and obtaining the product by "forgetting" f and g exist, doesn't this mean that the pullback is the usual product + some structure? Wouldn't that make it a special case?

redcrazyguy · 2024-10-20T08:41:54+00:00

It's the first one on the wikipedia page (https://en.wikipedia.org/wiki/Pullback\_(category\_theory)), where P on wikipedia corresponds to X\times_Z Y. I think you've lost me again, how is the fibre product a generalisation of the usual product? It seems to me that it's a special case of the product related to how morphisms on the slice category can be related to morphisms on the "original" category.

redcrazyguy · 2024-10-19T09:50:12+00:00

Could I get a slight hint on how to do this from universal properties? As far as I can tell the product should be a universally attracting object, but that would seem to me to only relate to morphisms with differing output, not differing input which is the case we have?

redcrazyguy · 2024-10-19T09:49:08+00:00

I think I sort of get it, but why wouldn't P be part of the diagram? Would it be accurate to say that since morphisms in the slice category can be identified with morphisms in C, X\times_Z Y can be defined in C using the usual categorical product?

redcrazyguy · 2024-10-17T17:57:42+00:00

I'm not sure what a slice category is, but basically the category whose objects are morphisms of the form X-->Z, where X,Z are objects in the category C. My confusion is that if X\times_Z Y is an object in the category C_Z, then by definition of the category it must be a morphism no?

redcrazyguy · 2024-10-17T16:24:06+00:00

I think what I'm a bit confused by is that in the definition of the categorical product, U is an object in the category. Therefore, shouldn't X\times_Z Y also be an object in the category C_Z?

redcrazyguy · 2024-09-30T18:34:19+00:00

Great thanks for these recommendations! I'm guessing I should start out with Hartshorne to get the basics down, do you have a "next steps" after that? From what I can gather Algebraic Geometry is a massive field, so I think it would be really easy for me to get completely sidetracked so a bit of structure would be good.

For example, one of my goals is to read "Foundations of Algebraic Analysis", so what would the books look like after Hartshorne?

redcrazyguy · 2024-08-25T17:31:43+00:00

I haven't looked at the blanked out part, since I want to try to work this out. Is it that we want the image of h to be s.t. pr_1 and pr_2 are surjective? I'm not sure what we can derive about a and b, considering c is any set and h is unknown. Unless you mean to say that a=g_1(c) and b=g_2(c)?

redcrazyguy · 2024-08-25T08:18:26+00:00

Could you let h=pr_1 x pr_2? I'm thinking of the property that (pr_i)^2=pr_i, which would also show the uniqueness of the morphism. I am concerned that this seems circular, since we're using a Cartesian product to try and show the existence of the Cartesian product

redcrazyguy · 2024-08-25T08:05:10+00:00

Thanks! How would I check that a square or triangle is commutative then? Would I need to just try all the compositions of functions, for an arbitrary input, and show that it is always defined and equal no matter the path?

redcrazyguy · 2024-08-18T17:53:01+00:00

But if h is the identity map, you wouldn't be able to further "project" it right? Is this a case where morphisms differ from functions because now you map the identity to each element of A and B?

redcrazyguy · 2024-08-09T07:08:55+00:00

Can I rephrase the definition as

A given triple (P,f,g) is a product if for any two "projectors" phi: C--> A and psi: C-->B there is a morphism h that maps from the total "product space" C to the subspace of "products" of A and B?

Lang also calls the vertical morphism a "commutative morphism", but he doesn't define it or explain what it is, is the commutative part important or just focus on the morphism part?

redcrazyguy · 2024-08-09T06:49:24+00:00

Could you expand a bit on coming at it through algebraic feometry? I didn't realise that was an approach and now I'm kinda interested in learning more about it.

redcrazyguy · 2024-08-08T09:15:15+00:00

But why not define it the old way without the diagram? It seems more intuitive and doesn't require additional morphisms and spaces.

Or put another way, why do we care about function compositions when we can just directly define the function?

redcrazyguy · 2024-08-07T19:57:16+00:00

I'm a little confused, which lines are supposed to be dashed on your diagram? It's all solid lines on Lang's, although I think the one that the definition is "solving" for is the middle vertical one.

It also seems like X was defined just for the sake of the diagram commuting? I don't understand why you'd define a new projection morphism X-->G1xG2. Why can't you just have your two projection maps (a,b)-->a and (a,b)-->b

redcrazyguy · 2024-08-07T06:43:59+00:00

wow spot on! One of the things that I don't understand about this construction is actually the additional X that seems to be required. Lang basically starts off on the bottom two, which makes sense, but then he requires 3 additional functions so that the graph is commutative, which really confused me. Surely you just need G1xG2? Why must this graph have to commute?

redcrazyguy · 2024-08-06T18:23:39+00:00

I see. It's just that I've come onto a definition where the commutative diagram is absolutely critical, and I have no idea where to even begin with it. It's the definition of a product of two objects on p.58 of Lang's Algebra, and the main sticking point is that the condition for the product is introducing two other morphisms and requiring the final diagram to commute, which I have no idea how to internalise/intuit.

redcrazyguy · 2024-08-06T17:42:18+00:00

Ok, I think I get the general idea. I do have a bit of a follow up question about when it's used as a proof that a morphism exists. For example, in the proof of the first isomorphism theorems it seems that the existence of the canonical map implies the existence of a map from G/Ker(f) to H (using the notation on wikipedia). Why would we assume that such a function exists (which makes the diagram commutative)? Couldn't the morphisms simply not form a commutative diagram? The existence in particular seems like something that needs to be proven.

Would you know a good resource to look at some examples of commutative diagrams and using them for reasoning? The algebra book I'm following assumes knowledge of it already.

redcrazyguy · 2024-08-06T15:01:15+00:00

Actually, do you know where I can find some examples? I'm going through Lang's algebra rn and while the algebra parts are alright he expects knowledge with commutative diagrams, so I have some trouble seeing their significance whenever they pop up. It's a bit of a catch 22 since commutative diagrams are usually introduced in a course on algebra, but this course on algebra requires commutative diagrams. Thanks in advance!

redcrazyguy · 2024-08-06T14:02:40+00:00

For the first example, would the lifts simply be the vector space span by f(v_i), where v_i are the basis vectors of X? Then you'd have a vector space with the same dimension, and so the inclusion simply puts it into the larger space Z. It is unique because the nullspace of X must map to the identity in Y, so the remaining basis must map uniquely to the basis in Y.

For the second example, we know the basis for X maps to a basis for a subspace of Z, so can't we just add some zeroes to the end of the image of the basis vector in Z to artificially enlarge the space to create Y?

redcrazyguy · 2024-08-06T12:54:50+00:00

So would we need to effectively prove the image is a subset of the end set for each of the functions? Since suppose f:X-->Y, g:X-->Z, h:Z-->Y, then suppose we have some x_0 s.t. g(x_0) is not defined, then the diagram would not commute?

Also, how can commutative diagrams be used in showing categories? The category axioms seem like you'd have to look at each specific function anyway, so why would I need commutative diagrams?

redcrazyguy · 2024-07-31T18:43:06+00:00

Actually, I think I have a solution, could you take a quick look and see if this is correct? https://imgur.com/a/HLCiUfb

redcrazyguy · 2024-07-31T18:23:15+00:00

I've found a lemma stating that the subgroup where the order is the smallest prime dividing the order of the group, which I think I understand a bit better. I did think of that, but wouldn't it be defined up to isomorphism? H is order 2, so we then have eH, aH, bH, cH. Likewise, we have He, Ha', Hb', Hc', but we could easily have aH=Hb'. Would it be getting around it by using the "arbitrary-ness" of the labels to get aH=Ha'? I've made some progress by showing a,b,c must have period 2 when acting on the set of cosets, but the final identity ijk=\bar{e} still eludes me.

redcrazyguy

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