New answers tagged

3

In this category, I find counting to be effective at disproving conjectures. If it did have a left adjoint $F$, then there would be a bijection $$ \hom(FA, B) \cong \hom(A, X \times B) $$ so we would have $$ b^{|FA|} = (|X| b)^{|A|} $$ for every finite cardinal $b$. Supposing $|X| > 0$, we would need $$ |FA| = |A| (1 + \log_b |X|) $$ for all ...


-1

This is only difficult if you get confused by the following facts that are stupidly written in all the textbooks A limit of a particular diagram is a right Kan extension A functor assigning limits to diagrams is a right adjoint but a left Kan extension. What happens if you apply a right adjoint to a limit is that you still have a limit, but if you apply ...


0

(I misread your question and wrote the answer about limits; but everything below holds with colimiting cocones in place of limiting cones). Given a functor category $[\mathcal E,\mathcal C]$, a diagram $\mathcal D\xrightarrow{J}[\mathcal E,\mathcal C]$, and a limiting cone $F\overset{\phi_A}\Rightarrow JA$ over $J$, it is not in general true that for any ...


3

I see that this question has already some good answer. Nonetheless allow me to give a personal perspective on the matter. The definition of currying and partial application are the following: currying is an operation that takes a function of two (or maybe more argument) and return a function-valued function partial application is an operation that takes a ...


4

No, it doesn't unless $X$ is a singleton. The very first condition to check for a functor to have a left adjoint is that it should preserve limits (such as products, equalizers...). But clearly in general, if $X$ has at least two element, $$X \times (Y \times Z) \not\cong (X \times Y) \times (X \times Z),$$ and so the functor doesn't preserve products, thus ...


3

Good question. No, it does not have a left adjoint. One of the most important properties of adjoints is that right adjoints commute with limits (including products and the terminal object), while left adjoints commute with colimits (including coproducts and the initial object). For example, since $\emptyset$ is an initial object and $X \times -$ is a left ...


1

Not really if you are considering pure category theory (as the tag suggests): $$\operatorname{Hom}(X \times Y, Z) \cong \operatorname{Hom}(X, \operatorname{Hom}(Y,Z)).$$ As usual when you have a canonical natural isomorphism, in most context you can replace one object with the other without changing anything. The notions of "returning" and "evaluating" and ...


-1

As far as I can tell, the only real difference that is ever noted is that partial application combines two steps into one: (1) first it curries a function $f$ and then (2) evaluates the resulting curried function $\tilde{f}$ with a given value. But then, at least in my unimportant opinion, that shouldn't be called "partial application", since one would ...


1

The answer to your question is kind of: strongest and weakest topologies are certainly terminal and initial objects in certain categories, but they are required to satisfy an additional property. What I'm writing is partially taken from the Joy of Cats which is freely available online, and partially taken from the nlab. Explicitly, the category of ...


2

Yes, the final topology is the final object in the category of topologies on the given set making the given functions continuous, which is the poset whose objects are such topologies and whose maps are inclusions of sets, and dually for the initial topology. These words weren't necessarily invented to agree with each other, though. Arbitrary limits and ...


0

Well I found that the problem is with the definition of $\kappa$-accessibility and many resources just don't care about this notational problem and take it for granted! Here we go: A $\mathtt {Set}$-functor $T:\mathtt {Set} \to \mathtt {Set}$ is defined to be $\kappa$-accessible for a regular cardinal $\kappa$ iff for all sets $X$ and all $x\in TX$ there ...


0

The category of elements of $\mathcal Set\xrightarrow{F}\mathcal C$ is also a comma category $(\{*\}\downarrow F)$, i.e. a kind of weak pullback of $\mathbf 1\xrightarrow{\{*\}}\mathcal Set\xleftarrow{F}\mathcal C$, in the sense that we have pair of functors $\mathbf 1\leftarrow(\{*\}\downarrow F)\xrightarrow{\Pi_F}\mathcal C$ together with a natural ...


2

It might be amusing to see how this works in the context of covering spaces. If $G$ is discrete, a $G$-bundle is specified by a representation $\rho: \pi_1(B) \to G$; this representation is determined by picking a basepoint in the total space $x \in E$, and sending a loop to the unique $g$ such that if you lift the loop to start at $x$, it ends at $gx$. This ...


4

To directly answer your question: Andrej meant "monomorphism". Since the objects of a topos don't typically consist of (raw, unstructured) sets, the usual notions of injectivity and surjectivity don't apply. But for convenience, people still talk about "injections" and "surjections" in those contexts (and mean monomorphisms and epimorphisms, respectively). ...


1

I have missed that $A$ and $Z$ are not objects of ${\sf C}_A$ but only objects of ${\sf C}$, hence there is no circularity in the definition.


2

For (1), you are right. For (2): if $S \to B$ is a subobject, and $f: A \to B$ is any morphism, then we can form (because $\mathscr B$ is finitely complete) the pullback $$ \require{AMScd} \begin{CD} f^{-1}(S) @>>> S\\ @VVV @VVV \\ A @>{f}>> B. \end{CD} $$ Because monomorphisms are stable under pullback, $f^{-1}(S) \to A$ is a subobject. ...


5

You trivialize $E\times_B E\to E$ using exactly the nullhomotopy of $f\circ p:E\to BG$. For this you need to know that principal $G$-bundles over $X$, up to isomorphism, are in bijection with maps $X\to BG$, up to homotopy, and that given $f:X\to BG$ classifying $q:Z\to X$ and a map $g:Y\to X,f\circ g$ classifies $Z\times_X Y\to Y$. The first fact is proven ...


2

I'll write only maps that are actual functions to avoid confusion about contravariance. A map $f:Y\to P(X)$ corresponds to a map $f':X\to P(Y)$ by $f'(x)=\{y:x\in f(y)\}$. Then $f''(y)=\{x:y \in f'(x)\}=\{x:y\in \{\hat y: x\in f(\hat y)\}\}=\{x:x\in f(y)\}=f(y)$. Switching $X$ and $Y$, we see that the prime operation is indeed bijective. I don't understand ...


2

The point you're missing is that in a general topos, the internal logic is not necessary Boolean, so $\neg$ does not necessarily give complements. Topological spaces are a good source of examples for intuitionistic logic; in a lattice of open sets, $\neg$ gives the exterior of an open set. Usually, the complement of the set is not open, and thus does not ...


1

Regarding the second question, the center crops up in a different way: If $G$ is a group viewed as a category and $1_G$ is the identity functor, then $Z(G)$ is the group of natural transformations $1_G \to 1_G$. It can be useful to extend this to more general categories: for an arbitrary category $\mathbf{C}$, we can define $Z(\mathbf{C}) = \hom(1_\mathbf{...


3

There is a sense in which they always coincide if $\prod_{j\in J}$ is understood to be an internal product--that is, a right adjoint to the functor $J^*:\mathbf{C}/1\to\mathbf{C}/J$ given by pullback along the unique morphism $J\to 1$. The intuition for thinking of such a right adjoint as a product is that such a pullback is like taking an object in $\mathbf{...


2

What you're saying seems confused. First, you say from your previous question you have concluded $\mathrm{Hom}_{\mathbf{Set}}(J,X)=X^J\cong\prod_{j\in J}X$ – do you not see that this is exactly the product (really power) and exponential coinciding in $\mathbf{Set}$? Second, you say that in the category of topological spaces, the product (really power) and ...


1

$\require{AMScd}$I don't quite understand the question, but basically, I think you're trying to define a particular double category. Objects. Sets Arrows. Relations Proarrows. Relations Squares. We assume that each square has at most one filler, and that it has a filler iff the condition $$\quad(a,a')\in\alpha\wedge(b,b')\in\beta\implies\...


2

Such a characterisation of 'relations between relations' is not possible. As a counterexample, let $A,B,A',B',\alpha,\beta$ be whatever you want them to be, let $R = \varnothing$, let $R' = A' \times B'$. Then: $(a',b') \in R'$ for all $a' \in A'$ and $b' \in B'$, so the statement on the right-hand side of your $\Leftrightarrow$ symbol is true for all $a,...


3

Well, it is quite trivial: both groups consist of a single object, so the functor can only map the first object to the second. For two arrows $g,h$ in the first group, and a functor $f$ to the second group, functoriality means that $f(gh) = f(g)f(h)$. But this is precisely the definition of a group homomorphism.


1

Interesting questions. Let start be saying that your observations on the subcategories of groups are correct: if you exclude the empty category all the subcategories of a group are exactly the submonoids of the group (by the way there is a subcategory that is full: the group itself). About the posets, a subcategory of a poset-category $P$ is a poset $Q$ ...


2

Jean Goubeault-Larrecq has written a detailed textbook-level introduction to "Lawvere" metric spaces from a topological point of view in Chapter 6 of his book Non-Hausdorff Topology and Domain Theory. He defines a hemi-metric space to be a set $X$ equipped with a set-function $d\colon X\times X\to[0,\infty]$ such that $d(x,x)=0$ $d(x,y)\leq d(x,z)+d(z,y)$ ...


1

I'll try to answer your questions in order. I think you're right, it should be enough to prove that $K(f+g)=K(f)+K(g)$, which is exactly the definition of additivity. Again I think that your argument is sound. He's being redundant because every semigroup-homomorphism between groups preserves identities and inverses. $K$ has been defined as the pointwise ...


4

This is another example of "equivalence instead of equality". Monic maps "ought" not be characterized by $f(x) = f(y) \implies x=y$: the right characterization ought to be $f(x) \cong f(y) \implies x \cong y$. (note: the following was written around the symmetric version; where we further assume $d(x,y) = d(y,x)$. Without this assumption the details are ...


1

No, because a map of abelian groups can preserve addition without preserving the identity and inverses, but this is not the case for subtraction. Thus you answer this yourself in 2: he's being redundant. A morphism of additive functors is an arbitrary natural transformation. Thus since $K$ is an equalizer with respect to natural transformations out of any ...


5

I assume you are looking for some real world examples of pseudo-metric spaces (that correspond to what you call Lawvere-metric spaces). Well a family of examples is given by the $\mathcal L^p$-spaces (where $p$ is real number in $[1,\infty]$). In general for $(\Omega,\mathcal F, \mu)$ a measured space, we have the vector space $\mathcal L^p(\Omega)$ whose ...


3

The answer is yes, both formulations are equivalent. Let's see the correspondence. Let's see that giving a contravariant functor $F:C^{op}\to D$ is equivalent to giving another functor $G:C\to D^{op}$. Assume we know $F$. At the level of objects it's clear what to do: for any $a\in Ob(C)=Ob(C^{op})$ we define $G(a):=F(a)\in Ob(D^{op})=Ob(D)$. Now, given ...


3

The point is that the $f_{ij}$ are the entries in a matrix: $$\begin{pmatrix} f_{11}& f_{12}\\ f_{21} &f_{22} \end{pmatrix}.$$ In fact, in the case when our category is the category of (left) $R$-modules for some ring $R$, this description tells us that a map $R^n\rightarrow R^m$ (where we think of $R^n$ as $R\oplus\dots\oplus R$ with $n$ copies of ...


2

If by all the $\times$ you mean $\times_k$, then your diagram is not necessarily cartesian, as the example of $X = Y = \text{Spec}(\mathbb{R})$ and $L = \mathbb{C}$ shows (use that $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C} \times \mathbb{C}$), so I assume the $\times$ on the right (resp. left) denotes fiber product over $k$ (resp. $L$). ...


2

For (1), you're right, but one can (and probably the author intended this) even strengthen this by saying that any four morphisms $f_{ij}: A_i\to B_j$ give rise to a morphism $f: A_1\oplus A_2\to B_1\oplus B_2$. For (2)... you're also right, it's just that what you say stands in no contradiction with the statement in question: namely, when you consider $p_i\...


3

The 'corresponds' just means that there is a correspondence between e.g. the class $\text{ Ob}(C)$ of objects of $C$ and the class $[1,C]$ of all functors from $1$ to $C$. Likewise for the other cases. In fact, you probably have already seen a similar statement in the special case where the category $C$ has only identity morphisms: elements of a class ...


1

Hint: functions don't need to be surjective. A functor $C \to D$ does not have to "represent" every object in $D$: there is a functor that maps a field $K$ to the abelian group $K^*$ of invertible elements of $K$, but not every abelian group is the group of invertible elements of a field.


1

Well, yes, one functor $1\to C$ determines exactly one object in $C$, but this can be any object. So, the functors $1\to C$ correspond to objects of $C$.


3

An Alexandrov space is canonically preordered by $x\leq y$ when $y$ is in all the opens $x$ is in. This is clearly reflexive and transitive, but in general need not be antisymmetric: the latter holds exactly for $T_0$ spaces. A preordered set gives rise to an Alexandrov topology with open sets the upward closed sets. It's not too hard to see this ...


4

The category of affine schemes is nothing but the opposite category of commutative rings. So $\mathrm{Spec}\, \mathbb Z$ is a terminal object in the category of affine schemes because $\mathbb Z$ is an initial object in the category of commutative rings (the only morphism $\mathbb Z \to R$ is the one sending $1$ to $1_R$...).


1

Even with the extra condition that $\mathcal{A}$ is a Grothendieck category, it may still have no simple objects. I think the following is the easiest example I know. Let $R$ be a (necessarily non-noetherian) commutative local ring with non-zero maximal ideal $\mathfrak{m}$ satisfying $\mathfrak{m}^2=\mathfrak{m}$. Let $\mathcal{C}$ be the category of $R$-...


1

The answer to your question is yes $f$ is indeed the zero of the abelian group $\mathscr C(A,B)$. Here is the proof. Since $\mathscr C(0,0)$ has only one object it is the trivial group (the group with only the identity). We know that in a pre-additive category composition is bilinear so if $r \in \mathscr C(A,0)$ and $l \in \mathscr C(0,B)$ (the only ...


2

Yes: Deconstruct your $f:A \to B$ as $gh$, where $h:A \to 0$ and $g:0 \to B$. As there is a unique map $g:0 \to B$, we must have $g+g = g$ and hence, by bilinearity, $gh + gh = (g+g)h = gh$. But then $$f + f = gh + gh = gh = f,$$ and cancelling $f$ from both sides yields that $f$ is the zero object in the Hom-set.


2

I think Lectures and Exercises on Functional Analysis by A. Y. Helemskii might be exactly what you're looking for. Quoting from the introduction: Perhaps the main idea is that our book is written from the categorical point of view. Everywhere we stress and comment on the categorical nature of the fundamental constructions and results (like the ...


4

From the comments, the original issue is solved and now you ask how to figure out the nature of this unique map. You can think of the map $X \to \operatorname{Spec} \mathbb Z$ as a map, which sends a point to the characteristic of its residue field.


2

One possible way to "see" composition of arrows of the original category $\mathcal{C}$ in $\operatorname{Ar}(\mathcal{C})$ is the following: Suppose $f:x\to y$ and $g:y\to z$, let $1_x,1_y,1_z\in\operatorname{Ar}(\mathcal{C})$ be the identity morphisms of $x,y,z$ respectively, then $f$ gives a morphism $1_f:1_x\to1_y$ in the obvious way, and similarly for $...


2

The arrow category is equipped with two functors $s, t : \text{Ar}(C) \to C$ giving the source and target of an arrow. Composition is a functor $$\text{Ar}(C) \times_{\text{Ob}(C)} \text{Ar}(C) \to \text{Ar}(C)$$ where the LHS is a (2-)pullback, with one of the maps $\text{Ar}(C) \to \text{Ob}(C)$ being source and one being target. This expresses precisely ...


4

Note that there exists a $G$-equivariant map $f:G/X\to G/Y$ iff some conjugate of $X$ is contained in $Y$, and there is a $G$-equivariant bijection iff some conjugate of $X$ is equal to $Y$. So the question is, if $gXg^{-1}\subseteq Y$ and $hXh^{-1}\supseteq Y$ for some $g,h\in G$, must $X$ be conjugate to $Y$? The answer is no. For instance, fix any ...


1

A map $\Phi:S\rightarrow \prod_{j\in J} E_j$ is the same thing as a family of maps $(\Phi_j:S\rightarrow E_j)_{j\in J}$, where $\Phi_j$ is just the $j$-th coordinate in $\prod_{j\in J} E_j$. So to give a map $f_0:S\rightarrow\prod_{i\in I}\prod_{\varphi\in M_i} G_{i,\phi}$, it is enough to give each "coordinates". For the coordinate corresponding to $(i,\...


1

An answer among others. A topology on $X$ is obtained by the relation $r\subseteq X\times \mathcal P(X)$ where $(a,S)\in r\iff a\in \overline S$. Then a "canonical morphism" $(X,r)\to (Y,s)$ is a function $f:X\to Y$ such that $x\in \overline{f^{-1}(T)}\implies f(x)\in\overline T$, which means continuity. But of course it is possible to chose other ...



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