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4

How about the category of sets that are not singletons? Monomorphisms, epimorphisms and binary products are easily seen to be the same as in the category of all sets, but the map $f:\{x,y\}\to\{a,b\}$ with $f(x)=a=f(y)$ is extremal "epi".


4

Let me give you another point of view on the five lemma. If you have a commutative diagram of $R$-modules $$ \require{AMScd} \begin{CD} 0 @>>> A @>>> B @>>> C @>>> D @>>> E @>>> 0 \\ @. @VfVV @VgVV @VhVV @VjVV @VkVV \\ 0 @>>> A' @>>> B' @>>> C' @>>> D' @>>> E' ...


3

Epimorphisms in the category of sets and partial functions are simply partial surjections (i.e. partial functions $f\colon X\to Y$, such that $f(X)=Y$). So $f_1$ is an epimorphism but $f_2$ is not. Let $f\colon X\to Y$ be a partial function, such that $f$ is not a partial surjection, i.e. there exists such $y_0\in Y$, that $f^{-1}(y_0)=\varnothing$. Then ...


3

I think you're mixing up two things: a monoid in a monoidal category, and the usual monoid viewed as category. Let $\mathbf C$ be a monoidal category with multiplication $⊗$ and unit $I$. A monoid in $\mathbf C$ is an object $M$ of $\mathbf C$ together with a multiplication morphism $m : M ⊗ M → M$ and a unit morphism $e : I → M$, which satisfy certain ...


3

Recall that a cone from $N$ to $F$ is a natural transformation from the constant functor $\Delta N$ to $F$. A cone is universal if it represents the functor $[I,C](\Delta -,F) : C^\text{op} \to \textbf{Set}$, i.e. if it induces isomorphisms $$C(X,N) \cong [I,C](\Delta X,F)$$ for all objects $X$ of $C$. Note that there is a natural isomorphism $$[I,C](\Delta ...


3

Define a partial order on the objects of $\mathcal{C}$ by declaring that $X \le Y$ if there exists a monomorphism $X \to Y$, i.e. if $X$ can be viewed as a subobject of $Y$. Then the initial object is the least element of $\operatorname{ob}\mathcal{C}$, and an object satisfies your condition exactly when it is an atom for this partial order. So I guess you ...


3

As you say, you can't do anything like this with an arbitrary adjunction, but the examples you mention are very special cases: they are monadic over $\mathrm{Set}$, and in that case such a construction always works. Basically you use the fact that every object $a ∈ \mathcal A$ is the coequalizer of maps $εFU, FUε : FUFUa → FUa$, so you'll want to take $ν$ to ...


3

$\DeclareMathOperator{\colim}{colim}\newcommand{\cat}{\mathbf}\DeclareMathOperator{\Hom}{Hom}$ For any locally small category $\mathbf C$, you can construct its cocompletion $\hat{\mathbf C}$ as the following locally small category: The class of objects is the class of small diagrams in $\mathbf C$ (i.e. functors from a small category to a small ...


2

Judging from your comment, you mean "spanning" (somewhat) in the sense of graph theory. More common than "subcategory spanned by the objects with property $A$" is the term "full subcategory of objects with property $A$"; at the very least I think this what is meant here: This is simply the subcategory that you get by keeping just the objects with property ...


2

No, of course not. Consider the case with Russell's socks. It is consistent that there is a countable family of pairwise disjoint sets of size $2$, whose union does not have any countably infinite subset. Namely, the coproduct $\coprod_{n\in\Bbb N}P_n$ is not isomorphic, or even comparable, with $\Bbb N$. But on the other hand, $\Bbb N$ can certainly be ...


2

Let $\mathcal{A}$ be a locally small category with a terminal object. Then $\mathbf{Fam} (\mathcal{A})$ is also locally small with a terminal object $1$, so we can define $\Gamma = \mathrm{Hom}(1, -) : \mathbf{Fam} (\mathcal{A}) \to \mathbf{Set}$. This functor has a left adjoint $\Delta : \mathbf{Set} \to \mathbf{Fam} (\mathcal{A})$, defined on objects by $X ...


2

A morphism $f: P \to Q$ is just a $G$-equivariant fiber bundle map (i.e. if $p_1:P \to B$ and $p_2: Q \to B$ are the bundles, then $p_1 = p_2 \circ f$). It turns out that all principal $G$ bundle morphisms are isomorphisms. See here for an elementary proof. Thus, in particular, $f_* : \pi_1(P) \to \pi_1(Q)$ is an isomorphism. Is this the sense in which $f$ ...


1

There is no way to turn a real vector space into a complex one in general. Think, for example, about vector spaces of odd dimension. The even dimensional ones can be turned into complex ones but in many ways, and there is no way to do it functorially. This means that what you wrote does not define a functor (nor anything else, reallythe, as the ...


1

Not really an answer, but I wanted to post this here in case anyone else ends up thinking about this thing. It might help set you on the right track. Anyway, after browsing through this collection of slides (see "Ingredients of Construction" slide) I now know that the morphisms in this category can be described using walled Brauer diagrams (see e.g. page ...



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