Tag Info

5

For question 2, $F\otimes_A -$ has a left adjoint iff $F$ is finitely generated, and the left adjoint is always exact. For if $(f_i)\in F^I$ is an element of an infinite product of copies of $F$, then it is easy to see $(f_i)$ is in the image of the canonical map $F\otimes_A A^I\to F^I$ iff $\{f_i\}$ is contained in a finitely generated submodule of $F$. ...

5

$f\circ g$, being an automorphism, has an inverse $h$. Then $(h\circ f)\circ g$ is the identity, so $g$ has a left inverse and hence is a split monomorphism. Similarly, $g\circ f$ has an inverse $h'$, so $g\circ (f\circ h')$ is the identity, hence $g$ has a right inverse, hence is a split epimorphism. Thus $g$ is an isomorphism. Obviously this is a symmetric ...

4

As Zhen Lin says in the comments, wiki is wrong. The exactness of the sequence $R\overset r{\underset s\rightrightarrows} X\xrightarrow f Y$ is equivalent to exactness - in the abelian sense - of the sequence $0\to R\xrightarrow{\begin{pmatrix}r\\s\end{pmatrix}}X\oplus X\xrightarrow{(f,-f)} Y\to 0$. You can prove $(P,r,s)$ is the kernel pair of $f$ iff ...

4

I like [Hirschhorn, Model categories and their localisations]. First, a word of warning: the book is divided into two parts, but the first part depends logically on the second part. Thus, beginners should start at Chapter 7, not Chapter 1. The great thing about [Hirschhorn] is that all the definitions and results are stated carefully and clearly, so it ...

3

Let me remove most of your assumptions and work with an arbitrary module $M$ over an arbitrary commutative ring $R$. We'd like to know when the functor $M \otimes_R (-)$ has a left adjoint. The answer is iff $M$ is finitely presented projective, in which case the left adjoint is $M^{\ast} \otimes_R (-)$ where $M^{\ast} = \text{Hom}_R(M, R)$. You can ...

3

John Baez, Aaron Lauda, Higher-Dimensional Algebra V, arXiv Dan Mardsen, Category theory using string diagrams, arXiv Peter Selinger, A survey of graphical languages for monoidal categories, arXiv

2

Take your favourite example of a universal arrow, say $\{a, b\} → U\{a, b\}^*$ (the free monoid generated by letters a, b). Now $\{a, b\}^*$ is certainly big enough to factor every arrow $\{a, b\} → UM$, and it will remain so even if you add some clutter, making it into eg. $\{a, b, c\}^*$. But now the factorization is not unique, because you can map $c$ to ...

2

Not exactly an solution to your question, but frankly it's too big to be fitting a comment. The distinction as to exactly what is being linear is very important. It is possible to approximate non-linear behaviours if we are allowed to make the linear space (as in linear algebra) large enough. Approximating solutions to smaller dimensional non-linear ...

2

Bounded quantification is saying "for all elements $x \in X$, it holds that ..." or "there exists an element $x \in X$ such that ...". Unbounded quantification is saying "for all (sets) $X$, it holds that ..." or saying "there exists a set $X$ such that ...". For some parts of mathematics, bounded quantifications entirely suffice. For instance, when talking ...

2

If $C=B/f(A)$ is not torsion, let $T$ be its torsion subgroup and use the quotient map $B\to C/T$ to show $f$ is not an epimorphism. Conversely, if $g_0,g_1:B\to D$ are such that $g_0f=g_1f$ and $D$ is torsion-free, $(g_0-g_1)f=0$, so $g_0-g_1$ factors through $B/f(A)$. Now use the fact that any map from a torsion group to a torsion-free group is $0$.

2

Your question is imprecise: at several points you mix up objects and their underlying sets. Here is a precise version: Suppose $(C, F)$ is a concrete category, that is, a pair consisting of a category $C$ and a faithful functor $F : C \to \text{Set}$ (the "underlying set" functor). Further suppose there is an object $c \in C$ such that $F(c)$ is the ...

2

A set with one element is indeed a poset. Any one-element set $A = \{a\}$, with the partial order $R = \{(a,a)\}$ (i.e. $a \leq a$) is a partially ordered set. Then for any poset $P$ there exists a unique map \begin{align} !:~&P\longrightarrow \{a\} \\ &p \longmapsto ~~a \end{align} which is indeed monotone. Thus a one-element set is a terminal ...

2

If $n>1$, show that multiplication by $n$ is a bimorphism $\mathbb{Z}\to\mathbb{Z}$ but not an isomorphism. (The hard part of this is showing it is epic; for this, you want to show that if $F$ is a free abelian group, then a homomorphism $\mathbb{Z}\to F$ is uniquely determined by where it sends $n$.)

2

There is a variant of ETCS with replacement, due to Colin McLarty, but I don't think this is really the right approach. For instance, for your specific problem of iterating the power object functor, the real issue is not replacement but rather that the induction principle you get out of the universal property of an NNO is too weak. Here's a non-parametrised ...

2

The "naive" coproduct of functors where you define $(\coprod_i F_i)(T)=\coprod_i F_i(T)$ in not a sheaf in general. If you sheafify, (at least in the case you're interested in) you should get a description like this: an element of $(\overline{\coprod_i F_i})(T)$ (the sheafification) is a decomposition $T_i$ of $T$ as a disjoint union $T=\coprod_i T_i$ (same ...

1

For the first question, I interpret it as asking why $F(Q,Vect_K)$ is a category. As Tobias Kildetoft mentioned in his comment, this is a general construction, namely, for any pair of categories $A$ and $B$, with $A$ a small category, we can define the functor category $F(A,B)$: its objects are functors $F:A\to B$ for any pair of objects $F$ and $G$, ...

1

You can simply say that if $A,B$ are two finite abelian groups, then $\operatorname{Hom}(A,B)$ is a finite set. However, the functor $A\mapsto\prod_{i\in\mathbb{N}}\operatorname{Hom}(A,\mathbb{Z}/p\mathbb{Z})$ does not always take a finite group to a finite set. Hence, it cannot be represented by a finite group. (An object representing this functor is by ...

1

Consider the composition $[a,b]\overset{G}{\rightarrow}[Ga,Gb]\overset{\varphi ^{-1}}{\rightarrow}[FGa,b]$. By Yoneda, it is determined by $(\varphi ^{-1}\circ G)(id _a)=\varepsilon _a$ and since, $\varphi ^{-1}$ carries epis to epis, and $G(id_a)$ is epic, $\varepsilon _a$ is an epi. But then, using the fact that $f:b\rightarrow a$ is monic (epic) ...

1

Given that the direct sum of finitely generated modules is finitely generated, the only thing that could possibly go wrong is that there is some homomorphism between finitely generated modules that does not have a finitely generated kernel. I leave it to you to translate this to a condition on $R$...

1

Your last definition is (basically) as general as it gets (so far as I know). Definition. Let $S$ denote a semiring (with both $0$ and $1$.) Suppose $X$ and $Y$ are modules over $S$. Consider a function $f : X \rightarrow Y.$ Then: $f$ is linear iff for all $x,x' \in X$ and $s,s' \in S$, we have $f(sx+s'x') = sf(x)+s'f(x')$. $f$ is affine iff ...

1

First of all, the result you're trying to prove isn't true as stated. It is only true if you restrict $\mathcal{C}$ to be the category of $k$-linear colimit-preserving functors. Second, you don't need a natural map $M\otimes_k G(R)\to M$; you need a natural map $M\otimes_k G(R)\to G(M)$. To construct such a map, first consider the case $M=R$. In that ...

Only top voted, non community-wiki answers of a minimum length are eligible