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2

A beautiful combinatorial answer was found by Gjergji Zajmi. For an expanded version, see the solution to Exercise 2 in PRIMES 2015 reading project problem set #1. (Ignore Section 0.1; it has nothing to do with this.) Unlike Martin's answer, this one directly gives a formula for $\dbinom{xy}{n}$ as a nonnegative linear combination of products ...


0

Let me actually give a full answer, as this is an important point in category theory. Proof of the fact that $E$ is a functor: For every arrow $h:D\rightarrow D^{\prime}$ in $\mathbf{D}$, you have defined $E\left( h\right) $ as the unique arrow $h^{\prime}:E\left( D\right) \rightarrow E\left( D^{\prime}\right) $ with $F\left( h^{\prime}\right) ...


1

According to G.M. Kelly's Examples of Non-monadic Structures on Categories (page 63), downloadable here, $\alpha$ is not necessarily an isomorphism if we only have a closed category with an adjoint to $[-,-]$. Unfortunately, as far I can tell, Kelly does not give a counterexample. Regardless, if we require the adjoint to be "internal" in the sense that we ...


0

I myself have been recently looking for a citation and a proof, but here (I think) is the idea behind it. First, to clear up the question, I will assume you are referencing the brief tidbit mentioned in the commutative diagram page on Wikipedia under "verifying commutativity" (https://en.m.wikipedia.org/wiki/Commutative_diagram). This means you are talking ...


6

Although there's a distinction between the noun and adjective abbreviations of monomorphism and epimorphism (mono vs. monic and epi vs. epic), it's fairly common to use iso for both the noun and adjective abbreviation of isomorphism. For example, both of these seem normal to me: $\mathcal{C}$ is a balanced category if every morphism $f$ of $\mathcal{C}$ ...


1

Let $F\colon\mathcal{C}\to\mathcal{D}$ and $G\colon\mathcal{D}\to\mathcal{C}$. be (covariant) functors. Then $G$ is a left adjoint to $F$ if there exists a natural isomorphism $$ \hom_{\mathcal{C}}(G(X),Y) \xrightarrow{\sim} \hom_{\mathcal{D}}(X,F(Y)) $$ For contravariant functors the same definition applies, after having made them covariant. There are two ...


6

There are two possibilities. One is the "naïve" or "strict" notion. A strict $G$-action on a category $\mathcal{C}$ is a group homomorphism $G \to \mathrm{Aut} (\mathcal{C})$, where $\mathrm{Aut} (\mathcal{C})$ is the group of automorphisms of $\mathcal{C}$. The fixed points of a strict $G$-action on $\mathcal{C}$ are defined in the obvious way and form a ...


1

Yes. It is straight forward to check the group axioms; they are inherited from the $G_i$. This construction can be found in every book which treats direct limits (which is the wrong name for directed colimits) and of course Wikipedia.


5

The identity $\binom{x+y}{n} = \sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k}$ is well-known, it is called the Vandermonde identity. The answer for $\binom{xy}{n}$ can be explained using the notion of a $\lambda$-ring, where here we consider the binomial ring $\mathbb{Z}$ with $\lambda^n(x)=\binom{x}{n}$. The main theorem on symmetric polynomials enables us to ...


2

The smallness condition is not especially important. The key condition is indeed idempotent-completeness. Suppose $A$ is a compact object in $\mathbf{Ind} (\mathcal{C})$. By definition, there is a filtered diagram $Y : \mathcal{J} \to \mathcal{C}$ such that $A$ is the colimit of $Y$ in $\mathbf{Ind} (\mathcal{C})$. But the canonical comparison map ...


4

If $\mathcal{C},\mathcal{D}$ are categories, then the projection functor $\mathcal{C} \times \mathcal{D} \to \mathcal{C}$ (which "forgets" the second coordinate) is not faithful (unless $\mathcal{D}$ is thin or $\mathcal{C}$ is empty).


4

One example is the forgetful functor from Schemes to Sets. Given two fields $k_1, k_2$, there may be many different field homomorphisms $k_1 \to k_2$ which give rise to many different morphisms $\text{Spec }k_2 \to \text{Spec }k_1$. However, the underlying sets of both of these schemes are single points, so there is a unique map between them in the category ...


2

The category with small coproducts freely generated by a category $\mathcal{C}$ is $\mathbf{Fam} (\mathcal{C})$, the category of families of objects in $\mathcal{C}$. The quickest description is as follows: $\mathbf{Fam} (\mathcal{C})$ is the category obtained by applying the Grothendieck construction to the (contravariant) functor $X \mapsto \mathcal{C}^X$. ...


1

David Spivak's Category Theory for the Sciences has lots of great examples (with solutions depending on your edition) that comes about as close to such a text as you're likely to find. It rides a fine line between mathematics and engineering/sciences with a stronger foot in mathematics. Keep in mind that there are multiple versions of his "open" text ...


2

If $C$ is a category with finite limits, you can make sense of a category object internal to $C$. You can also make sense of a groupoid object internal to $C$; just add a morphism which plays the role of the inversion. A group object internal to $C$ is then a groupoid object $G$ such that $\mathrm{Ob}(G)=1$, the terminal object. Isn't this covered by 1.? ...


2

Your method is not going to work unless you're careful about the choice of the projective modules $P$ and $Q$ through which $gf-\text{id}_M$ and $fg-\text{id}_N$ factor. Indeed, you could replace $P$ by $P\oplus P'$ for any projective module $P'$, changing the isomorphism type of $M\oplus P$. Instead, use the fact that $gf-\text{id}_M$ factor through a ...


0

Yes, that's correct. This is just the algebra homomorphism $\mathrm{Sym}_A(M) \to \mathrm{Sym}_A(0)$ induced by the module homomorphism $M \to 0$. You can also write down the addition on $E/X$, this corresponds to the algebra homomorphism $\mathrm{Sym}_A(M) \to \mathrm{Sym}_A(M) \otimes_A \mathrm{Sym}_A(M) = \mathrm{Sym}_A(M \oplus M)$ which is induced by ...


2

Suppose given a parallel pair $X \rightrightarrows Y$ in $\mathbf{Sh}$. Let $Y \to \tilde{Z}$ be the coequaliser in $\mathbf{Psh}$ and let $Z$ be the sheaf associated with $\tilde{Z}$; then $Y \to Z$ is the coequaliser in $\mathbf{Sh}$. Now, consider a morphism $Z' \to Z$ and define $X' = Z' \times_Z X$, $Y' = Z' \times_Z Y$, and $\tilde{Z}' = Z' \times_Z ...


1

A simple answer that covers most of the purely algebraic examples is that any class of structures that can be axiomatised by Horn clauses gives rise to a concrete category whose forgetful functor commutes with products. (In another question Concrete category with non-standard products, I use the term concrete category with standard products for this ...


2

Suppose a diagram of simply connected pointed spaces has limit $X$ in $\mathbf{Top}_*$. If it has limit $\tilde{X}$ in $\mathbf{Top}_1$ then there is a canonical map $\tilde{X}\to X$ that is the universal map from a simply connected pointed space to $X$. The universal cover of $X$ (when it exists) comes close to this, but the failure of lifting theorems for ...


1

Another way to get examples is to use non-standard forgetful functors. For example, $\mathsf{Set}$ is concrete over itself via the functor $X \mapsto X \amalg X$. But this functor doesn't preserve finite products.


1

In the book Rings with generalized identities by Beidar, Martindale and Mikhalev, Section 1.4 (from page 24 onwards) you can find the definition from commutative diagrams, the characterization with proof (which is the one given by Chindea Filip in his answer) and several properties. The main purpose of the coproduct in this book is to serve as the "home" ...


4

Historically, category theory was created to encode the functoriality of homology. Given a natural number $n$, to any topological space $X$, you can associate an abelian group denoted $\mathrm H_n(X)$ which gives you some kind of information on the space $X$. As such, it does not seems like an overwhelming construction. What matters in that construction is ...


2

Yes, there are two typos/inconsistent notations in the diagrams (as you found in your questions 3 and 4). Well, yes, the terminal object plays the role of the 'one-element set' in this abstraction, and a map $1\to X$ plays the role of 'selecting an element' of $X$. The map $e$ wants to select the identity element. You are right that one can also consider ...


3

Ok, I believe that the book try to do the following thing (although avoiding a lot of details). As a general case for every category $\mathbf C$ you can build a category $\text{Fam}(\mathbf C)$ whose: objects are families of objects in $\mathbf C$, that is stuff like $(c_i)_{i \in I}$ where each $c_i$ is in $\mathbf C$ morphisms from $(c_i)_{i \in I}$ to ...


7

What you are describing is the category of $\mathbb {Rel}$ of sets and relations. The objects of this category are sets and the morphisms are binary relations. This comes from the fact that there is a bijection between the sets of function of the form $X \to \mathcal P(Y)$ and (relations) subsets of the cartesian product $X \times Y$: this bijection is ...


3

If $R$ is a commutative ring, then the category of commutative bialgebras (resp. Hopf algebras) over $R$ is dual to the category of affine monoid (resp. group) schemes over $R$. This may be seen as an extension of the duality between commutative algebras over $R$ and affine schemes over $R$. However, I don't know any geometric description of coalgebras over ...


2

There are two categories you might be thinking of. 1) The category Rel of sets and relations, with composition defined via: $z \in (P \circ Q)(x) \Leftrightarrow \exists y \in Q(x) \ z \in P(y)$ This category is described in Giorgio Mossa's answer. 2) The category Mult of sets and multivalued functions, with composition defined via: $z \in (P \circ Q)(x) ...


3

Given a $\mathbf{Pos}$-enriched category $\mathcal{C}$ and an object $Y$ in $\mathcal{C}$, a downward-closed subobject of $Y$ is a monomorphism $m : X \to Y$ in $\mathcal{C}$ such that, for every object $T$ in $\mathcal{C}$, the induced map $$\mathcal{C} (T, m) : \mathcal{C} (T, X) \to \mathcal{C} (T, Y)$$ is the embedding of a downward-closed subset. In ...


2

The question is ancient, but IMHO the most convincing answer is missing: The definition of "subobject in a category $\mathcal{C}$" is chosen in such a way that it generalize the notion of $k$-vector subspaces (when $\mathcal{C} = \mathsf{Vect}_k$), the notion of subsets (when $\mathcal{C} = \mathsf{Set}$), the notion of subrings (when $\mathcal{C} = ...


0

$\mathsf{QBRing}$ is not algebraic via the usual forgetful functor, because limits in $\mathsf{QBRing}$ are formed by taking the limit in $\mathsf{Ring}$ and then coreflecting into $\mathsf{QBRing}$, so the usual forgetful functor doesn't preserve limits. For instance, in $\mathsf{QBRing}$ the kernel pair of the quotient $\mathbb{Z} \to \mathbb{Z}/2$ is the ...


7

Just recall that functors take isomorphisms to isomorphisms. Thus $G(\psi): G(F(X)) \xrightarrow{\sim} G(Y)$, and so the composition $\varphi \circ G(\psi): G(F(X)) \xrightarrow{\sim} Z$ is the desired isomorphism.


1

Firstly, I assume that objects of $U_\infty$ are identifed with objects of $\bf S^0$, i.e. with sets, via the functor $A_\infty$. I feel some confusion here. The notation $A:=B$ usually means to define the symbol $A$ as $B$. Now the category $U_\infty$ seems to be given, though you want to override the homsets. Its resolution is that either we simply ...


3

The category of $C^*$-algebras is complete. The limit of a diagram $(A_i,\lVert - \rVert)_{i \in I}$ of $C^*$-algebras has as underlying $*$-algebra the $*$-subalgebra of the $*$-algebra $\prod_{i \in I} A_i$ whose elements $x=(x_i)_{i \in I}$ are subject to two conditions: First, the usual matching condition: For edges $i \to j$ the map $A_i \to A_j$ should ...


3

In addition to Henning Makholm's crisp and clear answer, you might find the opening six pages of my Notes on Category Theory helpful. They too give the example of a monoid as a category, but also give some other examples of categories where the arrows are not functions in any ordinary sense. Another important illustration is the case of a posets treated as a ...


2

You already know that a monoid $M$ is a set with a unit $e$ and a binary operation. More precisely, if $a,b,c\in M$ then $$a\circ b\in M$$ $$(a\circ b)\circ c=a\circ (b\circ c)$$ $$e\circ a=a\circ e=a$$ Now, take any category $C$ with one object, $c$. Since $C$ is a category, we need to say what the arrows are. That is, what the morphisms $c\rightarrow c$ ...


0

It is the underyling set function obtained from $\overline f$. In detail, we are given a set $A$, we obtain the free monoid $M(A)$ in the usual way: The $x\in M(A)$ are words of finite length; the multiplication is cancatenation and the identity is the empty word denoted by $e$. On the other hand, $M(A)$ is also just a set so we can forget about its ...


4

No, that's not the way to view it. In order to view a monoid as a category, you have a single object $\mathsf{Andreas}$, and each element of the monoid is one morphism $\mathsf{Andreas}\to\mathsf{Andreas}$ in the category. The monoid operation is the composition in the category. So for the integers, you don't have a morphism "add 1", but a morphism that is ...


3

On the first question: yes, the points of the Cauchy completion can be identified with their distance functions. The points of any metric space (satisfying the condition $d(x,y)=0 \implies x=y$) can be identified with their distance functions -- this is just an obvious fact for metric spaces; in enriched category theory it corresponds to the Yoneda lemma. ...


5

As Martin says in the comments, one way to rephrase the question is that you want an example of a concrete category $(C, U)$ where $C$ has binary products but the forgetful functor $U : C \to \text{Set}$ does not preserve them. In the most familiar examples $U$ not only preserves products but all limits because it has a left adjoint / is representable, so ...


3

Here is a method to produce examples. Two specific examples will follow. Let $\mathcal{A} \to \mathsf{Set}$ be a "standard" concrete category. For example, you may imagine an algebraic category. In particular, I want that $\mathcal{A}$ is complete. Let $\mathcal{C} \subseteq \mathcal{A}$ be a coreflective full subcategory. Then it is well-known that ...


1

Since you're interested in considering objects that have elements I'll assume we are working with $\mathbf{Sets}$ the category of sets and functions. From the function $g \colon A' \to A' \oplus B'$ and from $inr \colon B' \to A' \oplus B'$ you can obtain the function $\bar g \colon A' \oplus B' \to A' \oplus B'$. By definition this function is such that ...


1

Counterexample: Let $I$ be the category given by a single nontrivial arrow $f:I_0\to I_1$. Let $C$ be two disconnected copies of $I$, say $\iota_0,\iota_1:I\to C$ map to the separate copies of $I$. Let $C'=I$. Define $\varphi:C\to C'$ so that $\varphi\circ\iota_i=1_I$ ($i=0,1$). Then, there is no natural transformation between $\iota_0$ and $\iota_1$ in ...


0

I think we're supposed to define $F(X, f) = F(\text{id}_X \times f)$. Then $F(X, \text{id}_{X'})$ does equal identity. And $F(X, f\circ g) = F(\text{id}_X \times (f \circ g)) = F((\text{id}_X \times f) \circ (\text{id}_X \times g))$ and since $F$ is a functor: $$ \dots = F(X,f) \circ F(X,g) $$


1

Products in many categories are said to possess a universal mapping property- formally, a product of two objects (say $A$ and $B$), is another object (say $P$) along with two arrows: $p_1:P \to A$ and $p_2:P \to B$ such that if $C$ is any other object (in our category), along with any pair of arrows, $f_1:C \to A, f_2: C \to B$, then there exists a UNIQUE ...


2

The trick is that composition in the op category has a different meaning. Remember that a Category is defined by its objects, morphisms, and the composition of those morphisms. In the op category, we have the same objects and the same morphisms, but composition has been redefined. $\DeclareMathOperator{\Hom}{Hom} $In particular: suppose that we have ...


0

Maybe, you can consider $G_1=G$ and $G_2=G\oplus H$ Clearly, there are two canonical morphisms, the injection $i:G_1\rightarrow G_2$ and the projection $\pi:G_2\rightarrow G_1$ Now if $Aut$ is a functor, no matter it's contravariant or covariant, a contradiction easily follows if you can construct $G, H$ satisfying $Aut(G)$ has a generator $g_1$ of order ...


3

$\mathbb{C}$ is just a particular Hilbert space, and a vector in $H$ is the same thing as a morphism $\mathbb{C} \to H$, so you can talk about these using only the dagger category structure of $\text{Hilb}$, together with the distinguished object $\mathbb{C}$. One reason to distinguish $\mathbb{C}$ is that it's the identity for the tensor product of Hilbert ...


1

Use the long exact sequence for homology to obtain $$\cdots\longrightarrow H_{n+1}(C_\cdot)\longrightarrow H_n(A_\cdot)\longrightarrow H_n(B_\cdot) \longrightarrow H_n(C_\cdot) \longrightarrow H_{n-1}(A_\cdot)\longrightarrow \cdots$$ and conclude. For example $C_\cdot,B_\cdot$ acyclic gives exact sequences $0\longrightarrow H_n(A_\cdot) \longrightarrow ...


4

Here are two relevant theorems from Gillman and Jerison, Rings of Continuous Functions, Van Nostrand 1960: Theorem 8.3. Two realcompact spaces $X$ and $Y$ are homeomorphic if and only if $C(X)$ and $C(Y)$ are isomorphic as rings. Theorem 10.6. Let $t:C(Y) \to C(X)$ be a ring homomorphism with the property that $t\mathbf{1} = \mathbf{1}$. If $Y$ is ...



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