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Please excuse my bad ASCII art. I'll do a real pic as soon as I figure out how. -- Fropuff 06:13, 2004 Jul 10 (UTC)
(Ah, those were the days. Geometry guy 21:36, 16 May 2007 (UTC))
Following discussion didn't provide a solution; any other input? -- Baarslag 16:57, 2 April 2006 (UTC)
The canonical injections are not necessarily injective, so maybe the link is confusing. Are they always mono though? --
Baarslag
13:36, 1 April 2006 (UTC)
Some googling turned up this paper which says on page 19 that in an extensive category, coproduct injections are mono. This suggests not only that the result is not true in general, but gives us a criterion for when it does hold. Now I just have to figure out what an "extensive category" is. Then it should be easy to find a counterexample. - lethe talk + 19:14, 1 April 2006 (UTC)
If I'm not mistaken, the following is a counterexample. The coproduct of Z_m and Z_n (integers mod m and n, respectively) in the category of commutative rings is Z_g, where g = gcd(m,n). This is because the equations 1+...+1=0 (m times), which holds in Z_m, and 1+...+1=0 (n times), which holds in Z_n, together imply 1+...+1=0 (g times). Clearly the canonical injections are not injective. I don't think they are monic either by the following argument. Let Z[x] be the free commutative ring on one generator x. This is the ring of polynomials in one variable with integer coefficients. Assume m > 2g. Let g_1, g_2:Z[x] -> Z_m be the unique homomorphisms with g_1(x)=g, g_2(x)=2g. Then g_1 and g_2 are not equal, but f(g_1(x)) = f(g) = 0 and f(g_2(x)) = f(2g) = 0, so f o g_1 = f o g_2. Can someone please confirm? 67.255.11.12 ( talk) 03:16, 3 January 2014 (UTC)
Define the following category:
Two unequal arrows p, q: D → A, v: A → C, w: B → C. And define vp = vq. C is a coproduct and v, w the canonical injections, since it's the only limiting cocone on A and B, but v is not mono by definition. Is this correct? -- Baarslag 23:17, 2 April 2006 (UTC)
Why is the arrow corresponding to f dashed in the second diagram? Is this common notation? Randomblue ( talk) 16:01, 30 September 2008 (UTC)
I'm having some trouble understanding the Hom_C duality/opposite discussion. I'm trying to reconcile it with a statement from a book on cohomology (R. Bott, differential forms in algebraic topology, page 46) when talking about the non-finite case: "... comes from the fact that the dual of a direct sum is a direct product, but the dual of a direct product is not a direct sum." I can't figure out how to reconcile this statement with what this article states, and thus would like to see a worked example of this. FWIW, since this is cohomology, its about Abelian groups/vector spaces. linas ( talk) 15:16, 30 November 2011 (UTC)
Doesn't a category need to be enriched in Ab as well as having finite biproducts to be additive? Not clear on that-- Daviddwd ( talk) 22:47, 27 August 2014 (UTC)
The adjoint to the product (cartesian product or tensor product) is hom, or the exponential object in the category of sets. See the article on currying for a lurid exposition of this adjunction. Now, the coproduct is opposite of the product. Is there an adjoint, and what can I say about it, either in some small category or in general?
I mean, if you start talking about additive categories and cohomology, then you get lead to tensor-hom adjunction and ext and tor functors, where you can say some abstract stuff. But it seems very domain-specific to cohomology; what can one say, more generally, that makes sense, and is still somehow useful? 67.198.37.16 ( talk) 07:13, 22 July 2017 (UTC)
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This article needs interpretation of the coproduct notation as it appears in 'summation' article for summation sequences. Please Atlantis President ( talk) 12:29, 14 March 2019 (UTC)
The definition given on this page, and on the page for products, defines a product and a coproduct as being specific objects for which there exist projection/coprojection maps respectively, with universal properties of factorisation through the product/coproduct respectively. Other sources ask that the projection/coprojection maps are actually part of the data of the product/coproduct. Perhaps one could keep it how it is currently stated, and treat the other perspective about taking product/coproduct objects along with witnesses of their product/coproduct structure? — Preceding unsigned comment added by 128.250.0.120 ( talk) 22:21, 15 April 2023 (UTC)
Is Yuan's work truly enough for this article? Mousefountain ( talk) 02:10, 4 January 2024 (UTC)
This is the
talk page for discussing improvements to the
Coproduct article. This is not a forum for general discussion of the article's subject. |
Article policies
|
Find sources: Google ( books · news · scholar · free images · WP refs) · FENS · JSTOR · TWL |
![]() | This ![]() It is of interest to the following WikiProjects: | ||||||||||
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Please excuse my bad ASCII art. I'll do a real pic as soon as I figure out how. -- Fropuff 06:13, 2004 Jul 10 (UTC)
(Ah, those were the days. Geometry guy 21:36, 16 May 2007 (UTC))
Following discussion didn't provide a solution; any other input? -- Baarslag 16:57, 2 April 2006 (UTC)
The canonical injections are not necessarily injective, so maybe the link is confusing. Are they always mono though? --
Baarslag
13:36, 1 April 2006 (UTC)
Some googling turned up this paper which says on page 19 that in an extensive category, coproduct injections are mono. This suggests not only that the result is not true in general, but gives us a criterion for when it does hold. Now I just have to figure out what an "extensive category" is. Then it should be easy to find a counterexample. - lethe talk + 19:14, 1 April 2006 (UTC)
If I'm not mistaken, the following is a counterexample. The coproduct of Z_m and Z_n (integers mod m and n, respectively) in the category of commutative rings is Z_g, where g = gcd(m,n). This is because the equations 1+...+1=0 (m times), which holds in Z_m, and 1+...+1=0 (n times), which holds in Z_n, together imply 1+...+1=0 (g times). Clearly the canonical injections are not injective. I don't think they are monic either by the following argument. Let Z[x] be the free commutative ring on one generator x. This is the ring of polynomials in one variable with integer coefficients. Assume m > 2g. Let g_1, g_2:Z[x] -> Z_m be the unique homomorphisms with g_1(x)=g, g_2(x)=2g. Then g_1 and g_2 are not equal, but f(g_1(x)) = f(g) = 0 and f(g_2(x)) = f(2g) = 0, so f o g_1 = f o g_2. Can someone please confirm? 67.255.11.12 ( talk) 03:16, 3 January 2014 (UTC)
Define the following category:
Two unequal arrows p, q: D → A, v: A → C, w: B → C. And define vp = vq. C is a coproduct and v, w the canonical injections, since it's the only limiting cocone on A and B, but v is not mono by definition. Is this correct? -- Baarslag 23:17, 2 April 2006 (UTC)
Why is the arrow corresponding to f dashed in the second diagram? Is this common notation? Randomblue ( talk) 16:01, 30 September 2008 (UTC)
I'm having some trouble understanding the Hom_C duality/opposite discussion. I'm trying to reconcile it with a statement from a book on cohomology (R. Bott, differential forms in algebraic topology, page 46) when talking about the non-finite case: "... comes from the fact that the dual of a direct sum is a direct product, but the dual of a direct product is not a direct sum." I can't figure out how to reconcile this statement with what this article states, and thus would like to see a worked example of this. FWIW, since this is cohomology, its about Abelian groups/vector spaces. linas ( talk) 15:16, 30 November 2011 (UTC)
Doesn't a category need to be enriched in Ab as well as having finite biproducts to be additive? Not clear on that-- Daviddwd ( talk) 22:47, 27 August 2014 (UTC)
The adjoint to the product (cartesian product or tensor product) is hom, or the exponential object in the category of sets. See the article on currying for a lurid exposition of this adjunction. Now, the coproduct is opposite of the product. Is there an adjoint, and what can I say about it, either in some small category or in general?
I mean, if you start talking about additive categories and cohomology, then you get lead to tensor-hom adjunction and ext and tor functors, where you can say some abstract stuff. But it seems very domain-specific to cohomology; what can one say, more generally, that makes sense, and is still somehow useful? 67.198.37.16 ( talk) 07:13, 22 July 2017 (UTC)
Hello fellow Wikipedians,
I have just modified one external link on Coproduct. Please take a moment to review my edit. If you have any questions, or need the bot to ignore the links, or the page altogether, please visit this simple FaQ for additional information. I made the following changes:
When you have finished reviewing my changes, you may follow the instructions on the template below to fix any issues with the URLs.
This message was posted before February 2018.
After February 2018, "External links modified" talk page sections are no longer generated or monitored by InternetArchiveBot. No special action is required regarding these talk page notices, other than
regular verification using the archive tool instructions below. Editors
have permission to delete these "External links modified" talk page sections if they want to de-clutter talk pages, but see the
RfC before doing mass systematic removals. This message is updated dynamically through the template {{
source check}}
(last update: 5 June 2024).
Cheers.— InternetArchiveBot ( Report bug) 00:30, 13 August 2017 (UTC)
This article needs interpretation of the coproduct notation as it appears in 'summation' article for summation sequences. Please Atlantis President ( talk) 12:29, 14 March 2019 (UTC)
The definition given on this page, and on the page for products, defines a product and a coproduct as being specific objects for which there exist projection/coprojection maps respectively, with universal properties of factorisation through the product/coproduct respectively. Other sources ask that the projection/coprojection maps are actually part of the data of the product/coproduct. Perhaps one could keep it how it is currently stated, and treat the other perspective about taking product/coproduct objects along with witnesses of their product/coproduct structure? — Preceding unsigned comment added by 128.250.0.120 ( talk) 22:21, 15 April 2023 (UTC)
Is Yuan's work truly enough for this article? Mousefountain ( talk) 02:10, 4 January 2024 (UTC)