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12

Here is a very simple example: let $P = \{a,b,c\}$ where $a \le b$ and $a \le c$ ($b$ and $c$ aren't comparable); let $Q = \{x,y,z\}$ where $x \le y \le z$. And let $f : P \to Q$ be defined by $a \mapsto x$, $b \mapsto y$ and $c \mapsto z$. $f$ is obviously bijective, and since $x \le y$ and $x \le z$ it's a poset homomorphism. But it's not an ...


8

(I don't know why the people are using again comments to answer the question; this way even "answered" questions will stay on the list of unanswered questions.) Yes, this is a category. Actually, every preordered set (a set with a reflexive and transitive relation $\leq$) can be regarded as a category in a natural way, where there is exactly one morphism $x ...


6

The problem is not in the morphisms … it is in the very definition of $\pi_1$ on objects: which group do you choose? There are, in principle, infinitely many groups $\pi_1(X,x)$ with $x\in X$, all different (because the underlying sets are different) but isomorphic. The isomorphism $\pi_1(X,x)\cong \pi_1(X,y)$ is not canonical, as it depends on the choice of ...


4

In abstract category theory morphisms are not required to be functions: we are given an arbitrary class, regarding its elements as arrows, equipped with domain and codomain information and an associative operation. If we want, we can specify so that for a funtion $f:A\to B$, let ${\rm dom\,}f:=B$ and ${\rm cod\,}f:=A$ and the operation is defined as ...


4

Let $C$ denote the category whose only object is $\Bbb R^n$ (for a fixed $n \in \Bbb N$) and whose morphisms are linear transformations. Let $D$ denote the category whose only object is the set $V = \{[x_1,\dots,x_n]^T:x_i \in \Bbb R\}$ and whose morphisms are matrices (composition is given by matrix multiplication). Given bases $A,B$ of $\Bbb R^n$ we ...


4

Yes, your interpretation is correct, but you better be careful about the size of your categories. Say you work with a universe $\mathbb U$, and you call small the elements of $\mathbb U$. Then, on one hand, the category of small topological spaces is a locally small category (meaning that the hom-sets are small). On the other hand, $\mathbf{Sheaves}$ is ...


4

One connection is that a pre-ordered set (i.e. a poset minus antisymmetry) can be viewed as a monad in the bicategory $Rel$ http://ncatlab.org/nlab/show/Rel


4

Yes. This is a corollary of the special adjoint functor theorem.


4

Every continuous functor $F : \mathsf{Set}\to \mathsf{Set}$ is isomorphic to $\hom(U,-)$ for some set $U$ and hence right adjoint to $- \times U$. Qiaochu has already mentioned that this follows from the Special Adjoint Functor Theorem. In the special case of $\mathsf{Set}$, one can make this more explicit as follows: Let $F : \mathsf{Set} \to \mathsf{Set}$ ...


3

Let me show how a family of isomorphisms $\mathcal{C}(X,Z)\cong\mathcal{C}(Y,Z)$ natural in $Z$ gives us an isomorphism $Y\cong X$. I will assume that you are familiar with functors, in particular the $\operatorname{Hom}$-funcotrs in question and natural transformations, because otherwise discussing implications of Yoneda's Lemma won't be very fruitful. For ...


3

If $G$ is a group, considered as a one-element category, then the natural isomorphisms from the identity functor to itself are exactly the automorphisms of $G$. If $X\to Y$ is any isomorphism in a category, then the induced natural transformation of functors $\operatorname{Hom}(-,X) \to \operatorname{Hom}(-,Y)$ is a natural isomorphism. You may have seen ...


3

I have a vague memory of being told that someone has proven that the Strom model structure is not cofibrantly generated. I would ask Boris Chorny or Carles Casacuberta. That said, a map $f \colon X \to Y$ is a Hurewicz fibration if and only if it lifts against the inclusion $Nf \to Nf \times I$, where $Nf$ denotes its mapping cocylinder. From this ...


3

The example that comes immediately to my mind is the obvious functor from a category $\mathcal{C}$ to its preorder-reflection $\mathcal{D}$, i.e. the category $\mathcal{D}$ whose objects are the same as in $\mathcal{C}$ and with a unique morphism $x \to y$ in $\mathcal{D}$ if and only if there is at least one morphism $x \to y$ in $\mathcal{C}$. It is full ...


3

1) Think of monoids as categories with a single object: any homomorphism $f : M \to N$ that is surjective but not an isomorphism is an example of a functor that is full but not faithful. 2) Pick a prime $p$ and let $\mathcal{P}$ be the category whose objects are the $\mathbb{F}_p^n$ and whose morphisms are polynomials in several variables. Let $\mathcal{F}$ ...


3

Note: the example given by Slade is correct, but there are more basic examples, like $(2\cdot) : \mathbb{Z} \to \mathbb{Z}$ The inclusion $\mathbb{Z} \to \mathbb{Q}$ is the usual counterexample of a ring epimorphism that isn't surjective Let $f : A \to B$. We say $f$ is epi if and only if for all $g,h:B \to C$ $g \circ f = h \circ f$ implies $g = h$ If ...


3

This is actually pretty easy to justify, but even easier to forget (as I did) or to never notice-as happens to readers of approximately every category theory book except Kashiwara-Schapira. Probably many students guess that $[\mathcal{C}^{op},\mathbf{Set}]$ is the opposite of $[\mathcal{C},\mathbf{Set}]$, even though this is badly wrong-I can think of at ...


3

Cat is a 2-category. So you can compose natural transformations as 2-cells. The counit of the composite adjoint equivalence will be $$F'G'GF \xrightarrow{1\gamma1} F'F \xrightarrow{\alpha} Id_B$$ and the unit will be defined in a similar way. This will work for adjoint functors as well. So in fact you get a category whose objects are categories and whose ...


3

$\pi_1$ preserves arbitrary products (not just finite ones); this is easy to check. $\pi_1$ does not preserve coproducts in general. See math:SE/320812. Seifert van Kampen's Theorem only applies under certain assumptions. $\pi_1$ does typically not preserve pushouts. For example $S^1$ is the pushout of two open intervals which have trivial $\pi_1$, but ...


3

In the construction of the colimit of sets, you have to define $\sim$ as the equivalence relation generated by the relation which you have defined. Similarly, in order to construct the colimit of algebraic structures of any type, you have to define $\sim$ as the smallest congruence relation generated by the relation (this ensures that the quotient will be ...


3

The two projection maps $G \times G \to G$ are homomorphisms, as is the map $G \to 1$. So is the diagonal map $G \to G \times G$. And are other complicated products and composites of these, like the map $G^3 \to G^6$ that, on a model, sends $(x,y,z) \mapsto (y,x,x,1,y,x)$.


3

Look at the definition. $U \subseteq Y$ is open if and only if $f^{-1}(U) \subseteq X$ is open. If $f$ isn't surjective (don't use the word projection), then let $y \in Y$ not in $f(X)$. Then $f^{-1}(\{y\}) = \varnothing \subseteq X$ which is open. Therefore you get the quotient topology on $f(X)$ and the discrete topology on its complement.


2

The correct statement is that an $f : X \times \cdots \times X \to X$ that is a homomorphism in the sense you described must either factor through $1$ or be a projection onto some factor. Think of the Lawvere theory of groups as the opposite of the category of free groups on finitely many generators. Then such an $f$ corresponds to a groups homomorphism ...


2

There is exactly one morphism from $X$ to $\emptyset$, namely the unique function from $\emptyset$ to $X$.


2

As usual, if you're confused about some general fact about categories, you should first see what it says about posets. In the context of posets products and coproducts correspond to two very natural operations, namely min and max. These are the two fundamental ways we might try to combine two elements of a poset to get another one, and roughly speaking the ...


2

Proving the tensor product of commutative $R$-algebras satisfies the universal property of the coproduct is straightforward, but it might be more illuminating to figure out the more general universal property of the tensor product of non-necessarily commutative $R$-algebras. Let $A$, $B$ and $C$ be $R$-algebras, not necessarily commutative. Let's compute ...


2

Martin Brandenburg has commented that an initial morphism is an initial object in a certain category. I want to point out that conversely, an initial object can be viewed as a special case of an initial morphism. Let $D$ be any category. Let $C$ be the terminal category consisting of a single object $X$ and a single arrow $\text{id}_X:X\to X$. Let $U:D\to ...


2

I think you are referring to the fact that a closure operator is an idempotent monad on a poset. http://ncatlab.org/nlab/show/closure+operator#definition Note that every poset can be viewed as a category with at most one arrow between each pair of objects. Your question would be better stated as "What is the connection between monads and closure operators?"


1

I'm assuming the question is: In the category $\mathbf{FinSet}$ what is the cardinality of $\DeclareMathOperator{Iso}{Iso}\Iso(X,X)$? We first have to understand what the isomorphisms are in $\mathbf{FinSet}$. It isn't too difficult to see these are exactly the bijections. So the cardinality of $\Iso(X,X)$ is exactly the number of bijections $X\to X$. This ...


1

An isomorphism from a set to itself is an automorphism. An automorphism is a permutation of the set since it is one-to-one and onto. Assuming the set $X$ is finite, ie. has $n$ elements, then the problem is reduced to how many ways can you rearrange $n$ elements, so the answer is $n!$.



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