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7

In general if you don't want to start with the monoidal structure, you start with a closed category. In a closed category $\mathsf C$, you have a bifunctor $$[-,-] : \mathsf C^{op} \times \mathsf C \to \mathsf C,$$ called the internal hom, and various other data that are somewhat "dual" to the axioms of a monoidal category (I put dual in quotes because this ...


5

Consider the pushout $$\begin{matrix} [0,1) &\to & [0,1] \\ \downarrow & & \downarrow \\ [0,1] &\to & X \end{matrix}$$ where all maps are inclusions. Here $X$ (an interval "with two $1$'s") is not Hausdorff.


4

Nice question! As user43208 says in comments, I presume you’re assuming $\newcommand{\CC}{\mathcal{C}}\CC$ is Cartesian closed throughout. The most straightforward way to see this as a natural transformation is to consider $Y$ as fixed. Then $Z^Y \times Y^X$ and $Z^X$ can each be seen as functors of two variables, contravariant in $X$ and covariant in ...


3

Provided the answer to Tobias' question is yes, i.e. you take a standard embedding ${\mathfrak g}{\mathfrak l}(2)\subset{\mathfrak g}{\mathfrak l}(3)$, Properties (1) and (2) are preserved, but (3) is in general not: First, note that under the assumption of Properties (1) and (2), Property (3) is equivalent to $W$ being finitely generated over the ...


3

I doubt we have strict equation between hom sets, rather we have $\hom(A,X)\cong\hom(B,X)$ naturally in $X$ (meaning a collection of bijections $\phi_X:\hom(A,X)\to\hom(B,X)$ which make the arising squares commute for each arrow $X\to Y$). To get an arrow $B\to A$, apply $\phi_A$ to $\def\id{\mathrm{id}} \id_A$, and to get an arrow $A\to B$, apply ...


3

The ring $\{ 0 \}$ admits a unique $k$-algebra structure, and it is easy to see that there is a unique $k$-algebra homomorphism $A \to \{ 0 \}$ for any $k$-algebra $A$. (If your definition of $k$-algebra does not allow for $\{ 0 \}$, then it is wrong.)


2

This is only a small addition to Najib's very nice answer which is nevertheless too long for a comment. It is even interesting to look at the structure of a closed category in the following very restricted setting: We consider only posets $({\mathsf P},\leq)$ considered as a categories. We consider only those closed structures in which the unit is a ...


2

The point is this: for any chain complex $A_{\bullet}$ in $\mathcal{A}$, we can form a a chain complex $(\operatorname{Cyl} A)_{\bullet}$ such that there is a natural bijection between morphisms $(\operatorname{Cyl} A)_{\bullet} \to B_{\bullet}$ and pairs of morphisms $A_{\bullet} \to B_{\bullet}$ with a chain homotopy between them. (See here.) More ...


2

The free product $G\ast H$ of two groups $G,H$ is defined and is the coproduct of $G$ and $H$ in the category of groups regardless of whether the groups are abelian or not. Of course, the free product $G\ast H$ a fortiori also satisfies the universal property of the coproduct when restricted to abelian target groups $A$, that is ...


2

As for your question, Galois comodules in that sense provide a useful framework to study Galois corings, which in turn are a framework to study Holf-Galois extensions, which in turn are a generalization of the theory of Galois extensions of rings to the situation in which the group is replaced by a Hopf algebra, which itself is a generalization of the good ...


2

Ok, the solution isn't that bad. Consider an infinite sequence of monomorphisms in $\cal C$ $$A=A_1\overset{i_1}{\hookrightarrow}A_2\overset{i_2}{\hookrightarrow}A_3\overset{i_3}{\hookrightarrow}\cdots\hookrightarrow\mathrm{colim}_n\,A_n=X$$ We need to prove that the inclusion $A\hookrightarrow X$ is still in $\cal C$, that is, that the inclusion ...


2

Perhaps I've misunderstood what you wrote, but it seems to me that your statement about the presheaf cokernel being a sheaf is false. E.g. take $X := \mathbb C^{\times}$ with its usual topology, let $\mathscr G := \mathscr O_X$ denote the sheaf of holomorphic functions on $X$, and let $\mathscr F$ denote the sheaf of locally constant functions on $X$ which ...


2

Welcome to the club! the club of (relatively) few people who realized that the standard textbook notation of a comma category morphism as an ordered pair ($(\phi,\psi)$ in your example) is incomplete/incorrect. To make a short story even shorter: the notation for comma category morphisms should be a quadruple, not a couple. For example - in your left ...


1

I am quite surprised that no one mentions here what I am about to mention, which it by the way the very definition of universal property, but anyway. I will give you the first example of universal property I saw when I was a kid that was coined as "universal property" to me, and then move to the explanation of what a universal property actually is. It was ...


1

Every functor $F : \mathcal A → \mathcal B$ between Abelian categories which reflects left (or dually: right) exactness must be faithful (if $f, g : A → B$ are mapped by $F$ to the same morphism, then $\mathrm{id} : FA → FA$ is the kernel of $Ff-Fg = 0$, but then $\mathrm{id} : A → A$ must be the kernel of $f-g$, which implies $f = g$). Now if $F$ is ...


1

Take for a simplest example $$R_1=\{(a,x),(a,y)\},\\ R_2=\{(x,z)\},\quad R_3=\{(y,z)\}\,.$$ Note also that we have equality with union instead of intersection.


1

They're called abstract sets. Personally, I would simply call this a "set," or perhaps a "mere set" or "unstructured set" to emphasize that there's no further structure around.


1

I only just saw this question. I'm a little surprised that none of the answers said concretely what $Set^{op}$ is. It's the category of complete atomic Boolean algebras, or equivalently complete Boolean algebras of the form $P(X)$ where $X$ is a set. The morphisms are morphisms of complete Boolean algebras (so functions that preserve arbitrary meets and ...


1

Suppose that $\mathsf{hom}(A, -)$ has a left adjoint $T_A$. In other words you have a natural (in $B$, $C$) isomorphism $$\mathsf{Ab}(A, \mathsf{hom}(B,C)) \cong \mathsf{Ab}(T_A(B), C).$$ Since $\mathsf{hom}(\mathbb{Z}, C) \cong C$, it follows that $T_A(\mathbb{Z}) \cong A$. As all left adjoint, $T_A$ preserves colimits, so for any family $\{\alpha\}$, ...


1

Well, neither for positive integers, addition is not always bigger than mulitplication: just think about $1\cdot 1\cdot 1 < 1+1+1$. So we shouldn't expect anything like $X\times Y \ge X+Y$ in general (whatever these symbols would mean). By a (group) presentation we mean a syntactic entity: a pair $\langle X,\Gamma\rangle$ where $X$ is a set (of ...



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