Questions tagged [universal-property]

For questions about universal properties of various mathematical constructions.

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Universality of correlators in $\beta$ Hermite ensembles

Recently I got interested in the world of Random Matrix Models and I bumped into some generalizations of the usual random matrix theories classified by Dyson whose probability density functions are: $...
Physicist in disguise's user avatar
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Universal Property of tensor product of $\mathbb Z_2$-graded algebras.

If $A$ and $B$ are two $k$ algebra's with a $\mathbb{Z}_2$-grading then I know that a $\mathbb{Z}_2$-graded structure can be defined on their tensor product. One does this by altering the ...
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Universal property of the completion of a first-countable abelian topological group

Let $G$ be a first-countable abelian topological group. Its completion $\hat{G}$, set-theoretically, is the family of Cauchy sequences in $G$ modulo the equivalence relation $(x_n)\sim (y_n)$ iff $x_n-...
Elías Guisado Villalgordo's user avatar
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What are colored morphisms/arrows intended to mean in these diagrams?

I've been reading more category theory as a prerequisite to understanding some more complicated theorems, and for the first time I'm running into arrows that are distinctly colored. Examples include ...
Nate's user avatar
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Looking for properties of a pseudo-logarithm where you add all the (prime factor$\cdot$exponent) terms in an integer's prime factorization?

If, for positive integer $n=p_1^{q_1}p_2^{q_2}\dots p_k^{q_k}$ with the right side the prime factorization of $n$, we define $J(n)$ as $p_1q_1+p_2q_2+\dots+p_kq_k$, does $J(n)$ have any generally &...
Lieutenant Zipp's user avatar
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Is this diagram of the free product of two groups correct?

From Wikipedia: "In mathematics, specifically group theory, the free product is an operation that takes two groups $G$ and $H$ and constructs a new group $G ∗ H$. The result contains both $G$ and ...
Nate's user avatar
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Wikipedia definition of Image in Category theory applied to Set

According to Wikipedia, the image of a morphism $f:A \to B$ is a monomorphism $m:I \to B$ for which there is some $e:A \to I$ with $f=m\circ e$, and such that the following universal property is ...
Diego Mathemagician's user avatar
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Objects of categories with no morphisms from them other than endomorphisms

Let $\mathcal{C}$ be a category. My question is the following. Is there a name for an object $\zeta \in \mathrm{ob}(\mathcal{C})$ such that, for every object $X \neq \zeta$, there exists no morphism $...
Tian Vlašić's user avatar
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Continuous maps between covering spaces

Here is the problem statement: Let $p:X\to S^1$, $q:Y\to S^1$ be covering maps, and $s_0=1\in S^1\subseteq \mathbb C$ is a basepoint for the unit circle. Let $M$ be the set of continuous maps $f:X\to ...
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Do commutative diagrams on projections lift to a commutative diagram on a product category?

I have a family of categories $\{A_i\}_I$, their product category $\Pi A_i$, a category $B$ and a family of functors $F_i:B\rightarrow A_i$. Then there is also a unique functor $F_I:B\rightarrow \Pi ...
sysyphusV's user avatar
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If $a^{p_1} b^{q_1}\ldots a^{p_n} b^{q_n} = e$ then $S = \{a,b\}$ is not a free generating set of $G = \langle S \rangle$

Let $S = \{a,b\}$ be a generating set for a group $G$. If a non-trivial word in $a, b \in S$ equals the identity $e$ of $G$, i.e., $$a^{p_1} b^{q_1} a^{p_2} b^{q_2} \ldots a^{p_n} b^{q_n} = e$$ for ...
stoic-santiago's user avatar
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Confusion on Exercise 3.6.ii in Riehl: showing the product is associative via a unique nat. isomorphism

This is an exercise (I believe) to practice methods using the functoriality of limits. The exercise is as follows: For any pair of objects $X,Y,Z$ in a category $\mathsf{C}$ with binary products, ...
William's user avatar
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The "characteristic property" of the tensor space

This relates to page 309 of John Lee's textbook on Smooth Manifolds. Specifically Proposition 12.7, the characteristic property of the tensor product space. I fundamentally fail to see how this would ...
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Associativity of tensor product using the universal property (and not elements)

My question I thought the author of this question was really pushing for what I'm asking here and I don't think it was ever fully addressed in other questions (happy to be wrong!). My main question is ...
cheyne's user avatar
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Are pseudometric completions canonical in any sense?

Let $P$ be a pseudometric space, and let $M_P$ be the metric space obtained by quotienting out points separated by zero distance. We can always complete $M_P$ to $\overline{M_P}$ by forming the ...
WillG's user avatar
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Show that $C^*(\mathbb Z)$ is the universal $C^*$-algebra generated by a unitary.

Show that $C^*(\mathbb Z)$ is the universal $C^*$-algebra generated by a unitary. I.e.,show that for any unital $C^*$-algebra $A$ containing a unitary $u$, there is a unique unital *-homomorphism $\...
analysis lover's user avatar
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Localising a ring twice gives ring isomorphic to localising by each subset in turn

Given a ring $R$, suppose we have two multiplicative subsets $S,U \subseteq R$ (which contain 1). Write $US = \{us \mid u \in U, s \in S\}$ also a multiplicative subset, containing both $S$ and $U$ ...
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How to derive this from the universal property of the ring of fractions?

(All the involved rings are commutative with identity.) The "universal" definition that I follow is this: Let $S$ be a multiplicative subset of a ring $A$ which also contains $1_A$. Then a ...
Atom's user avatar
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Is there any way to prove$\{v_i \otimes w_j\}$ (with$(i,j) \in I \times J)$are basis for tensor product without using universal property?

Let $V,W$ be a vector space over $K$ with basis $\{v_i\}(i\in I), \{w_j\}(w\in J)$ $\begin{eqnarray} S_1 &=& \{(v_1 + v_2,w)-(v_1,w)-(v_2,w)|v_1,v_2\in V,w\in W\}\\ S_2&=&\{(v,w_1+w_2)...
丸薬がんやく's user avatar
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Is this the correct formulation for the universal property and pushout diagram for cokernel?

The following question is taken from $\textit{Handbook of Mathematics}$ by Thierry Vialar,page 608 I consulted the Vialar's handbook and hoping to see what the dual formulation of cokernel in terms of ...
Seth's user avatar
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Universal property of multilinear maps (how to make them a tensor product)

context A tensor product of $V$ and $W$ is a couple $(T,h)$ where $T$ is a vector space and $h:V\oplus W\to T$ is a linear surjective map satisfying the following universal property : Let $f:V\oplus W\...
Laurent Claessens's user avatar
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1 answer
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Does the direct sum have a universal property in the category of groups?

In other categories, like modules, the direct sum is the coproduct, giving it a neat category theory description. The direct product of groups, which is very similar to it, is the product in the ...
Zoe Allen's user avatar
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In category theory, does a definition via a universal property induce a functor?

In category theory one frequently defines concepts via universal properties. Examples are (co)equalizers of pairs of morphisms having the same domain and codomain (co)kernels of morphisms in ...
Kritiker der Elche's user avatar
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1 answer
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Question about the meaning of a "region" in mathemathics.

I am currently going over double integral. One concept that seems to appear a lot is "region". I have consulted different resources and it seems like there is no objetive definition of what ...
xyz's user avatar
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Topical example of a universal arrow to and from a functor

If $ S\colon \mathsf D\to \mathsf C $ is a functor between categories and $ c\in \mathsf C $, Mac Lane in his Categories for the Working Mathematician defines a universal arrow from $ c $ to $ S $ as ...
GeometriaDifferenziale's user avatar
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AF-Algebras and Matrix Algebras

I have been having some trouble with an exercise from Murphy's "$C^*$-Algebras and Operator Theory" recently. Chapter 6 Exercise 6 is as follows: If $A$ is an AF-Algebra (i.e. a direct limit ...
LSK21's user avatar
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1 answer
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Embedding a free algebra in a product.

Let $A$ be an algebra; set $V=HSP(A)$. Then for any $X\neq\emptyset$, there is an embedding of the free algebra $F_{V}(X)$ into the algebra $A^{A^{X}}$, $\Psi\colon F_V(X)\rightarrow A^{A^X}$ given by ...
Mashiath Khan's user avatar
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Has this constructible property of structures in model theory been studied?

I was looking into model theory where a first-order language can be interpreted with a structure and a variable assignment. Here is an example of a FOL with interpretation: We have terms: Atomic ...
Kevin De Keyser's user avatar
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Uniqueness of the morphism in the universal property

If we want to define a product of objects in a category (cf. Wikipedia Page), we need to find a morphism and it must be unique. In our notes, the professor gives some examples such as the product of ...
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Definition of the evaluation map as a universal element of a representation

This is exercise Exercise 2.3.iii from Riehl's "Category Theory in Context": The set $B^A$ of functions from a set $A$ to a set $B$ represents the contravariant functor ${\rm Set}(-\times A,...
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1 answer
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Proving the existence of an homomorphism $k[x,x^{-1}] \to R$ where $R$ is a ring containing $k$ and a unit

I’d like to understand how to prove the following statement that I found in this answer: For any commutative ring $k$, the ring $k[x; 1/x]$ is characterized by the following "mapping property&...
lanero's user avatar
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1 answer
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The tensor product as a universal morphism

Given two vector spaces $V,W$ (say, over $\mathbb{C}$), we can form their tensor product $V \otimes W$, and we get a (bilinear) map $i : V \times W \rightarrow V \otimes W : (v,w) \mapsto v \otimes w$....
Sambo's user avatar
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Reference Request: Adjoint Functors and Universal Morphism

I currently work on non-category theoretical research but use some tools from category theory. I am not well-versed in category theory and also don't want to expand the preliminaries of my work too ...
justanotherhumanbeing's user avatar
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How to construct the map which verifies that the tensor product of algebras is the fibered coproduct using universal properties?

Let $A \rightarrow B$ and $A \rightarrow C$ be commutative ring morphisms. I am trying to verify the universal property of the fibered coproduct for the ring $B \otimes_A C$ with maps $i_B: B \...
rosecabbage's user avatar
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2 votes
2 answers
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Characterizations of product topology and box topology

Given a collection of topological spaces $X_i$ indexed by the elements $i$ of a set $I$, we consider the set product $P = \prod_{i \in I} X_i$ with projections $p_i : P \to X_i$. There are two methods ...
Paul Frost's user avatar
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Convenience of using universal property of tensor product

On the wikipedia page of universal property, it says one of the motivations of universal property is The concrete details of a given construction may be messy, but if the construction satisfies a ...
Tom's user avatar
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Direct limit characterized up to isomorphism

This is from Atiyah & Macdonald - Exercise 2.16: We are given a direct system $(M_i,\mu_{ij})$ with direct limit $M=\underset{\longrightarrow}{\lim} M_i$. Show that the direct limit is ...
iceberg56's user avatar
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Show that two tensor products are isomorphic.

Let $M$ and $N$ be two $R$-modules,then a pair $(T,\varphi)$ where $T$ is an $R$-module and $\varphi:M\times N\to T$ is a bilinear map is called a tensor product if for any pair $(P,f)$ where $P$ is ...
Kishalay Sarkar's user avatar
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Universal property of completion of a measure space

Let $ (X,\mathscr M,\mu) $ be a measure space. The completion of $ (X,\mathscr M,\mu) $ is the measure space $ (X,\bar{\mathscr M},\bar\mu) $ where $$ \bar{\mathscr M} = \{{E\cup F : \text{$ E\in \...
giovanniadeodato's user avatar
1 vote
1 answer
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If $R$ is an integral domain, then $Frac(R[x_1,…,x_n])= Frac(K[x_1,…,x_n])$ where $K=Frac{(R)}$

I’m trying to show that if $R$ is an integral domain, then $Frac(R[x_1,…,x_n])= Frac(K[x_1,…,x_n])$ where $K=Frac{(R)}$ (here $Frac{(R)}$ denotes “field of fractions of $R$”). I think I can use the ...
dahemar's user avatar
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Complexification of $O(n,\mathbb{R})$

The definition of complexification for a Lie group $G$ is a complex Lie group $G_C$ with a continuous homomorphism $\phi: G\to G_C$ with the universal property that, if $f: G → H$ is an arbitrary ...
l4teLearner's user avatar
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Commutative monoid "generating" algebraic number field

Let's say that we have an algebraic number field, which is also an ordered field (e.g. no roots of -1), and whose ring of integers is a UFD. We would like to form a multiplicative, commutative free ...
Mike Battaglia's user avatar
1 vote
1 answer
105 views

Dependence of strict transform on the subscheme along which we blow up

It is stated in https://stacks.math.columbia.edu/tag/080C that Note that taking the strict transform along a blowup depends on the closed subscheme used for the blowup (and not just on the morphism $...
SeparatedScheme's user avatar
3 votes
1 answer
73 views

Can a product map factor up to homotopy through the fat wedge

For $n \geq 2$, for $i = 1,\dots,n$, let $f_i \colon X_i \longrightarrow Y_i$ be maps of based $CW$-complexes, and consider a map $$f \colon \prod_{i=1}^n X_i \longrightarrow \prod_{i=1}^n Y_i$$ such ...
Matt's user avatar
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3 votes
1 answer
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Direct sum of injections is injection, categorically

Given $R$-module maps $\{f_i \colon M_i \to N_i\}_{i \in I}$, the universal properties of the direct sums $\oplus M_i$ and $\oplus N_i$ give a unique map $\oplus f_i \colon \oplus M_i \to \oplus N_i$ ...
Maxwell Jiang's user avatar
1 vote
1 answer
68 views

Proving the universal property for the localization functor $L_E$

I am trying to prove the following statement: If the functor $L_E$ exists, $(iii)$ for any map $g: X \to Y$ where $Y$ is $E_*$-local, there is a unique map $\tilde{g}: L_E X \to Y$ such that $\tilde{g}...
weird's user avatar
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3 votes
1 answer
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Formalizing the universal property of the tensor product

Introduction (a bit long) I am very new to category theory and algebra, in general. I recently saw the formalized concept of a universal morphism, here in WikipediA. I will repeat the definition with ...
Diego Mathemagician's user avatar
2 votes
0 answers
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Universal property of the (Vistoli-)sheafification

Given a presheaf $F$, in Notes on Grothendieck topologies, fibered categories and descent theory there is a construction of the sheafification (Proof for theorem 2.64). The first part is the ...
Muster Maxfrau's user avatar
1 vote
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Missing universal properties

Given a category $\mathcal{C}$, we can use copresheaves $H \in [\mathcal{C},Set]$ or presehaves $H \in [\mathcal{C}^{\mathrm{op}},Set]$ to state left or right universal properties. Existence of a ...
antoine_vm's user avatar
3 votes
2 answers
118 views

Functor demonstrating universal property of quotient spaces

The universal property of quotient spaces is, as quoted from wiki: The quotient space $X/\sim$ together with the quotient map $q:X\to X/\sim$ is characterized by the following universal property: if $...
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