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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 ...


0

I think the point is the triangulated structure on $K(\mathcal{A})$. One can obtain $D(\mathcal{A})$ directly by localising $Ch(\mathcal{A})$ with respect to the family of all quasi-isomorphisms, since the family of all quasi-isomorphisms in $Ch(\mathcal{A})$ satisfies the axioms for a closed multiplicative system. But then it's hard to find a triangulated ...


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 ...


0

The technical term for a set of values having no properties except that values are distinct is "set".


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.


0

A set is a collection of distinct elements in which order doesn't matter. A multi-set is a collection of elements (with possible repitition) in which order doesn't matter. Neither of these (or the next terms below) require or need any operations to be defined for them. For example, $\{ziggyswooglehorn, 5, \color{#C00}{\rm Red}\}$ is an example of a set ...


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 ...


1

Googling for equalizer family morphisms (also in Google Books and Google Scholar) returned some hits. It seems that some people use the name multiple equalizer. Adámek, Herrlich, Strecker: Abstract and Concrete Categories. The Joy of Cats mention this notion in Example 11.4(2), p. 194. Another book using this notion is Castellini: Categorical Closure ...


1

Yes, this is true. In an additive category matrix representations of maps between biproducts works. So the diagonal map for an object $X$ can be represented by the matrix $\left[\begin{smallmatrix} 1_X \\ 1_X \end{smallmatrix}\right]$. The inverse of this will be of the form $\left[\begin{smallmatrix} a & b \end{smallmatrix}\right]$ for some $a, b \in ...


1

The induced module will not be irreducible in general. For example, start with a $\mathfrak g$-dominant weight $\lambda$, viewed as a weight of $L$. Then there is a finite dimensional irreducible representation $V$ of $\mathfrak g$ with highest weight $\lambda$. If $W$ is the irreducible $L$-representation of highest weight $\lambda$, then there is a ...


3

If I understand correctly: Fix a category $\mathbf{C}$ and an object $\Omega$ in this category. A $\mathbf{J}$-shaped doodle to $\Omega$ is just a $\mathbf{J}$-shaped diagram in the coslice category $\mathbf{C} \downarrow \Omega$. A whatsit to a doodle $D$ is a cocone to $D$; hence the "greatest common factor" of $D$, being the initial whatsit to $D$, could ...


3

I suppose you mean injective objects with respect to the class of all monomorphisms. Recall that a monomorphism in $\mathbf{Top}$ is precisely an injective continuous map. Thus: Lemma. Every non-empty indiscrete space is injective. Proof. Let $Z$ be non-empty and indiscrete. Then every map $X \to Z$ is automatically continuous. Thus, given any ...


3

Good observation! Indeed it happens often that filtered colimits preserve finite limits, and as for a morphism $f: X\to Y$ being mono is equivalent to $id, id: X\rightrightarrows X$ being a pullback of $(f,f)$, in these situations monomorphisms are preserved. You find this imposed as a condition in topos theory, where a geometric morphism $f: {\mathcal ...


2

The result perhaps can be proved inductively showing that $\text{index}\,( T- \delta) = \text{index}\,T$ for $\delta$ of rank $1$. A useful particular case is showing by hand that $\text{index}( I - \delta) = 0$, starting with rank of $\delta=1$. The different cases presented suggests that there could be a uniform proof. We sketch one below. Break it into ...


5

You are not required to exhibit an inverse, only to prove existence. Thus, in the finite case, where the always-injective map to the second dual is actually a bijection, by dimension-counting, you can prove existence of an inverse without choosing anything: given $w$ in the second dual, there is unique $v\in V$ mapping to $w$. "Defined" $\alpha^{-1}(w)=v$. ...


7

The fact that $\alpha_V$ has an inverse is indeed less natural. As you know, $\alpha_V$ is a well defined natural embedding also when $V$ is infinite dimensional, but then it does not have an inverse usually. This implies that the construction of the inverse has to be somewhat less elegant than that of $\alpha_V$. However, choosing a basis does not ...


2

Yes, isomorphic: we can map the sup-lattice $(X,\le)$ to the inf-lattice $(X,\ge)$. This naturally extends on morphisms, and has a strict inverse.


0

Let $(\mathcal{A},\otimes)$ be a monoidal category. It is called abelian monoidal if $\mathcal{A}$ is an abelian category and $\otimes$ is an additive bifunctor (see here, Section 1.5). An example given the linked paper is the category of bimodules over a ring (in fact any closed abelian monoidal category can be exactly embedded in a bimodule category).


3

$\def\H{{\mathcal Hom}}\def\HH{{\operatorname{Hom}}}$Everything you wrote seems correct for the locally free case. Though its easy to overlook subtle things with these type of arguments, it looks good to me. For the second I believe there is a natural map $f^*\H_Y(E,F) \to \H_X(f^*E,f^*F)$. I will actually define a map $\H_Y(E,F) \to f_*\H_X(f^*E, f^*F)$ ...


4

If it can help you, in The Joy of Cats pag 121 there is this nice diagram:


2

Injection and surjection can be usefully defined at least in concrete categories. You can say that a morphism $f$ in a concrete category $(\mathcal C, U)$ is injective if $Uf$ is an injection. As the forgetful functor $U$ is assumed to be faithful, it in particular reflects monomorphisms, so $f$ injective implies $f$ is monic. Injectivity is certainly not ...


2

Yes, more generally, you have the following (which applies to your situation when setting ${\mathcal D} := {\mathcal C}^{\text{op}}$). Let ${\mathcal C},{\mathcal D}$ be finite length abelian categories and let ${\mathscr F}\dashv{\mathscr G}$ be an adjunction between the exact functors ${\mathscr F}:{\mathcal C}\rightleftarrows {\mathcal D}: {\mathscr ...


1

For general semisimple ${\mathfrak g}$, if ${\mathbb Z}\Phi$ is the root lattice, then any ${\mathfrak g}$-module $X$ in ${\mathcal O}$ decomposes as $X = \bigoplus\limits_{C\in {\mathfrak h}^{\ast}/{\mathbb Z}\Phi} X_C$, where $X_C := \bigoplus\limits_{\lambda\in C} X_\lambda$. As vector spaces, this is ok since it is only nested way of writing the weight ...


3

Remember that in category theory a kernel of $f\colon A \to B$ is not an object $K \subseteq A$ but a map $K \to A$. So the hypothesis that $\pi$ and $\varphi\pi$ have the same kernel means exactly that such a $j$ exists. If the arrow you're asking about existed then this would be a map that $\ker\varphi\pi$ factored through but $\ker\pi$ didn't, so they ...


2

This answer is also very late. There is a recent book by David Spivak titled "Category Theory for the Sciences." Its sole purpose is to connect the "real world" with category theory. You can find an older draft here from the authors homepage.


2

This too is a late answer, but in case anyone is still interested, here is a discussion with links about the use of category theory in biology/bioinformatics and genetics. Also, while not specifically a book on applications of category theory, the book Conceptual Mathematics by William Lawvere (an undergrad book, so not super advanced, but still a very nice ...


2

If $J$ is only locally small, the definition of $\mathrm{Lan}(f)(F)$ still makes sense even if $f^\ast$ is not definable without jumping to a bigger universe. It satisfies the 'local' adjointness property : there is a natural transformation $\eta \colon F \to \mathrm{Lan}(f)(F) \circ f$ universal in the sense that any $F \to G \circ f$ factors through ...


2

Here's one you might know. If $f: H \to G$ is a group homomorphism, then $f^*: [G,\mathsf{Vect}] \to [H,\mathsf{Vect}]$ is restriction of group representations, denoted $\operatorname{Res}_f$. The left adjoint $\operatorname{Ind}_f: [H,\mathsf{Vect}] \to [G,\mathsf{Vect}]$ is the induced representation functor. If $G$ and $H$ are finite, then ...


3

$A / \operatorname{Ker} f$ is called the coimage of $f$. There is always a canonical map $\operatorname{CoIm} f \to \operatorname{Im} f$, which can be defined in two dual ways (exercise!). An abelian category is often defined to be a preabelian category in which this canonical map is an isomorphism for every $f$. If instead you define an abelian category to ...


1

(3) is kind of open-ended. I have two things to say. If $A \overset{a''}{\rightarrowtail} A'$, then $A \overset{a'a''}{\rightarrowtail} B$ factors through $A' \overset{a'}{\rightarrowtail} B$, so you do have the original setup, it's just that the additional subobject $A \overset{a}{\rightarrowtail} B$ is extraneous information. If we put the previous point ...


3

Well, as far as I know, the only way to guarantee such an arrow is to induce it by some universal property. In this case, universal property of cokernel. Consider commutative diagram where $a = a'f$ and $\nu$, $\nu'$ are cokernels (observe that $a = a'f$ forces $f$ to be mono). We have $$\nu'a = \nu' a' f = 0$$ so there is unique map $g\colon B/A\to B/A'$ ...


2

You are right, it finally involves cokernels (=coimages of the next arrows). Your first line is correct: taking cokernel of both sides of $\def\im{\rm im\,} \im f=\ker g$, we'll get $\def\coker{\rm coker\,} \def\coim{\rm coim\,} \coker f=\coim g$. For the converse, take kernel of both sides. Then, for the objects, your second line should read $${\rm ...


-1

This doesn't give $\text{Im} f\cong \text{Coker} g$, it gives $\text{Im} f\cong \text{coker} f$, and modulo this mistake your argument is correct. Thus the decomposition is the one you've given.


2

You've got the right setup. The trick is that $\lim_{j \in J} \operatorname{colim}_{i \in I} I(Fj,i) = \lim_{j \in J} 1 = 1$ for any categories $I,J$, and functor $F: J^\mathrm{op} \to I$. The first equation is because the colimit of a representable functor is always 1 (exercise!). On the other hand, $\operatorname{colim}_{i \in I} \lim_{j \in J} I(Fj,i) = ...


3

The only real problem here is notation -- we're talking about functions whose values are functions, so we end up with chains like $f(x)(y)$ which is the function that $f$ returns at $x$, applied to $y$, and so forth. Once we're over that (conceptual) hurdle, we can simply write $$ f:(A^B)^C\to A^{B\times C} \qquad f(g)(b,c) = g(c)(b) $$ and then show that ...


2

Hint: $(A^B)^C$ is the set of maps from $C$ into $A^B$ (which itself is the set of maps from $B$ into $A$) and $A^{B\times C}$ is the set of maps from $B\times C$ into $A$. Define your bijection $\Phi: A^{B\times C}\rightarrow (A^B)^C$ by sending a map $f:B\times C\rightarrow A$ to the map $g: C\rightarrow A^B$ defined by $g: c\mapsto f(-,c)$ (where ...


11

Yes indeed, you just found the famous Yoneda Lemma (http://en.wikipedia.org/wiki/Yoneda_lemma). Concretely, while you only needed surjectivity of $f_{\ast}$ to deduce the existence of $g: B\to A$ with $fg=\text{id}_B$, you can use injectivity to deduce from $f_{\ast}(gf)=f(gf)=(fg)f=\text{id}_B\ f=f\ \text{id}_A=f_{\ast}(\text{id}_A)$ that $gf=\text{id}_A$ ...


2

No; even in the category of connected abelian Lie groups, tori are not injective objects. For ease of notation, we denote by $\mathbb{T}$ the torus $\mathbb{R}/\mathbb{Z}$. Here is a counterexample: Fix rationally independent, irrational numbers $\xi, \eta$. Let $f : \mathbb{R} \to \mathbb{T}^2$ be given by $$ f : x \mapsto (x\xi, x\eta). $$ Obviously $f$ ...


2

Here is one thing you can say, although it may not be the sort of result you have in mind. First of all, it does not suffice that your tensor category be abelian; if $M$ isn't finitely supported you also need infinite coproducts. It is better to work with symmetric monoidal cocomplete $k$-linear categories (this includes the condition that the monoidal ...


3

This is not possible in general. The problem is essentially whether left adjoints are faithful. For a counterexample to this, look at abelianization. Even if you restrict to tensor-hom adjunctions, it is not true in general. Take for example the two morphisms $a,b:\mathbb{Z}/(5)\rightarrow \mathbb{Z}/5$ such that $a(1)=1$ and $b(1)=-1$. These are equal ...


3

This functor is represented by the free idempotent, or $\mathbb{Z}[x]/(x^2 - x) \cong \mathbb{Z} \times \mathbb{Z}$. The statement that the corresponding functor to sets in fact lifts to Boolean rings says that $\text{Spec } \mathbb{Z} \times \mathbb{Z}$ naturally has the structure of a Boolean ring object in $\text{Aff}$. This structure comes from the ...


5

No, not every algebraic theory with finite biproducts is commutative. For instance, the category of $R$-modules for any ring $R$ has finite biproducts, but the theory of $R$-modules is commutative if and only if $R$ is commutative. But that's basically the only exception. Indeed: Theorem. Let $\mathbb{T}$ be a Lawvere theory. The following are ...


2

Given $\mathfrak{R}\longrightarrow f^{-1}\mathfrak{S}$ and $\mathfrak{S}\longrightarrow g^{-1}\mathfrak{T}$ gives you $\mathfrak{R}\longrightarrow f^{-1} \mathfrak{S}\longrightarrow f^{-1} (g^{-1}\mathfrak{T}) = (g\circ f)^{-1}\mathfrak{T}$, which is the wanted arrow. Of course, the arrow $f^{-1} \mathfrak{S}\longrightarrow f^{-1} (g^{-1}\mathfrak{T})$ is ...


1

The trick is to compare the arrows in $\mathbb{C}$ to those in $\mathbb{Set}^{ \mathbb{C}^{op} }$ or, if you prefer these terms, compare arrows in $\mathbb{C}$ to natural transformations in $\mathbb{Set}^{ \mathbb{C}^{op} }$. You can express this using homsets in the style of your answer but I will express it by diagrams. It would be worth working through ...


2

For groups this is not true. Let $F_n$ be the free group with $n$ generators, then $F_2$ has a subgroup isomorphic to $F_3$ and obviously $F_3$ has a subgroup isomorphic to $F_2$.


2

He means the two central squares, $\nu\theta=\lambda\zeta$ and $\xi\lambda=\mu\nu$.


1

It seems to me the answer is quite simple. On objects $P: C^{op} \to Set$ we have $F_!(P) = \int^{c \in C} P(c) \cdot F(c)$, where for a set $X$, the notation $X \cdot c$ indicates a coproduct of an $X$-indexed collection of copies of $c$. Let $\theta: P \to Q$ be a natural transformation. Then $$F_!(\theta) := \int^c P(c) \cdot F(c) \stackrel{\int^c ...


3

Kevin Carlson's answer is assuming you are working in a category with a zero object, but it actually works the same way in the full generality of kernel pairs. Suppose kernel pairs of $f$ and $q$ exist and let's denote them $k_1,k_2 \colon R \rightrightarrows A$ and $\ell_1,\ell_2 \colon S \rightrightarrows A$ respectively. Then, $fk_1=fk_2$ and $j$ is ...


3

Here’s a way to make your first idea work, though the pushout isn’t $\Bbb R/\Bbb Q$. Let $f:\Bbb Q\to\Bbb R:x\mapsto x$ and $g:\Bbb Q\to\Bbb R:x\mapsto -x$. The resulting pushout is the quotient of $\{0,1\}\times\Bbb R$ by the relation $\langle i,x\rangle\sim\langle j,y\rangle$ iff $\langle i,x\rangle=\langle j,y\rangle$, or $i\ne j$ and $x=-y\in\Bbb Q$. ...


3

These separation properties often fail to be preserved when something is not closed. When $X=Y=S^1$, you can take $A$ to be the open arc $A=\{(a_1,a_2)\in S^1\mid a_2>0\}$, the map $f:A\to X$ is the inclusion, and $g:A\to Y$ is the inclusion as well. So we basically just glue the two copies of $S^1$ along the arc $A$. The constructed space is not ...



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