24
votes
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Examples of a categories without products
For fields the most basic problem is fields of different characteristic. If $K$ and $L$ are fields, then the product $K\times L$ (if it existed) would have to be a field that can map into both $K$ ...
- 323k
20
votes
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Compact subset in colimit of spaces
As you suggest, choose a sequence of points $x_n\in K\cap X_n\setminus X_{n-1}$ (possibly replacing $(X_n)$ with a subsequence). Let $A=\{x_n\}$. Then if $B\subseteq A$, then $B\cap X_n$ is finite ...
- 323k
20
votes
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Calculating (co)limits of ringed spaces in $\mathbf{Top}$
The forgetful functor $\mathsf{LRS} \to \mathsf{RS}$ has a right adjoint. The right adjoint "$\mathrm{Spec}$" is a rather direct generalization of the spectrum of a commutative ring. You can find the ...
- 155k
14
votes
Examples of a categories without products
Let $K$ be a field. In the category of fields, let's prove $P = K \times K$ (categorical product) doesn't exist in general. If it did, it would come equipped with two projection morphisms $\pi_1, \...
- 6,236
14
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Accepted
Is every set a filtered colimit of finite sets?
The answer is yes: every set is the union of its finite subsets.
So take $I = P_{\text{finite}}(X)$ with as morphisms the inclusion maps, and $F : I \to \text{Set}$ the inclusion.
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12
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Examples of a categories without products
Posets (viewed as categories) provide many examples of categories without certain limits. A product of two elements in a poset is simply a greatest lower bound, so you just need to make a poset where ...
- 35.6k
12
votes
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Are fully faithful functors stable under pullback?
There is a bijective-on-objects/fully faithful orthogonal factorisation system on Cat. Hence, as a right class of a factorisation system, fully faithful functors are closed under pullback in Cat.
For ...
- 4,706
12
votes
Accepted
A category which direct limits but no general colimits
Consider the category with two objects and only identity arrows. Or more generally, any poset which has least upper bounds for all chains, but not arbitrary joins (like the disjoint union of two ...
- 72.1k
10
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Is there a coproduct in the category of path connected spaces?
They don't exist. For instance, suppose there existed a coproduct $Y=X\coprod X$, where $X$ is a point. Then there would be a unique map $f:Y\to [0,1]$ sending the first copy of $X$ to $0$ and the ...
- 323k
10
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Is every set a filtered colimit of finite sets?
One answer already mentions the diagram of finite subsets of $X$. You would have to check that taking the union of this system actually is the colimit (which is an easy exercise).
Since you asked for ...
- 12.4k
10
votes
Accepted
Can the fundamental group and homology of the line with two origins be computed as a direct limit?
Yes, $X$ is the direct limit of this sequence. This is essentially immediate from the universal property of quotient spaces: a map out of the direct limit is just a map out of $\mathbb{R}\sqcup \...
- 323k
10
votes
Accepted
Example of complete category with no initial object
None of the examples of categories with no initial object you describe are complete. It is actually tricky to write down an example here. The difficulty is that by the adjoint functor theorem, a ...
- 407k
10
votes
Accepted
Any criteria for a category to have all connected limits?
A category has all connected limits iff it has all pullbacks, equalizers, and filtered limits. First any limit can be built as a filtered limit of finite limits, and if the original limit was ...
- 323k
9
votes
Is a direct limit of Noetherian rings necessarily Noetherian?
In fact, every ring is a direct limit of Noetherian rings (here by "ring" I mean "commutative ring"). For any ring $R$, the set of finitely generated subrings of $R$ is a directed set under inclusion,...
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9
votes
Accepted
Why is $\mathbb{Z}[1/p]$ the direct limit of $\mathbb{Z}\xrightarrow{p}\mathbb{Z}\xrightarrow{p}\mathbb{Z}\to...$?
There's a better way to see why $\Bbb Z[1/p]$ is the direct limit (in my opinion): set up an isomorphism from the original direct limit to another one that's easier to understand:
$$\require{AMScd}
\...
- 150k
9
votes
Accepted
When do coproducts map canonically to products?
You have the right idea. The fact that $*$ is both initial and terminal (it's then called a zero object) implies that between any two objects $x$ and $y$ there is a unique zero morphism $0 \colon x \...
- 6,555
9
votes
Accepted
What is the intuition behind pushouts and pullbacks in category theory?
Pullbacks are fibred-products, i.e., a product with some compatibility restrictions. The terminology came from differential geometry when you really pull differential forms or their bundle on $B$ ...
- 31.3k
9
votes
Accepted
Is there a simple abstract reason why a profinite group is an inverse limit of finite groups?
As evidence that you cannot expect an "abstract nonsense" proof of this, the corresponding statement for Jónsson-Tarski algebras is false. A Jónsson-Tarski algebra is a set $X$ together ...
- 323k
8
votes
Accepted
Why does a union have to be disjoint to constitute a coproduct?
Try taking the coproduct of a set with one element with itself and use ordinary union. Pick two different maps into another set. Do they factor through one map from this potential coproduct, which has ...
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8
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Why does a union have to be disjoint to constitute a coproduct?
Matt Samuel has given you why specifically the union fails. Here is why the disjoint union has to be the correct notion of "union".
In category theory, when talking about sets, we don't care what ...
- 35.6k
8
votes
Accepted
End of Hom-profunctor in Grp
You're either approaching, or already made without saying so explicitly, the realization that the end of the hom bifunctor is the set of natural endomorphisms of the identity functor. This follows ...
- 51.4k
8
votes
Accepted
What exactly are the `size issues' preventing formation of presheaves being a left adjoint to some forgetful functor?
The problem is that you cannot choose a domain and codomain for such a putative adjunction consistently and simultaneously. The statement that we have is that the category of presheaves on $C$ is the ...
- 51.4k
8
votes
Accepted
Does the limit of a diagram with a single arrow exist?
If you're thinking about the diagram $\bullet \to \bullet$ (and identity arrows), then its limits are indeed uninteresting.
In fact, the limit of $A\to B$ is always just $A$.
The more general ...
- 42.9k
8
votes
What is the intuition behind pushouts and pullbacks in category theory?
Pullbacks generalise many common situations; they can be thought of as equationally defined sub-objects or as the subobjects of products that satisfy certain equations.
Here are a few examples of ...
- 2,112
8
votes
A category which direct limits but no general colimits
Consider any nontrivial group as a 1-object category. Then it has all filtered (co)limits (exercise: if all the morphisms in a filtered diagram are isomorphisms, then any object in the diagram is a (...
- 323k
8
votes
Accepted
Inverse limit of $\left(\mathbb{Z}/p^n\mathbb{Z}\right)_{n \in \mathbb{N}}$
This is false (I'll correct it on Wikipedia). The correct statement is that the direct limit / filtered colimit of $\mathbb{Z}/p^n \mathbb{Z}$ with these maps is the Prufer $p$-group $\mathbb{Z} \left[...
- 407k
8
votes
Accepted
The direct limit of roots of unity
Yes, send $a/b \in \mathbb{Q}/\mathbb{Z}$ to $\zeta_{b}^{a}$, where $\zeta_{b}$ is a primitive $b$-th root of unity.
- 14.7k
8
votes
What are the end and coend of Hom in Set?
The most important example of an end is $\int_{c : \mathcal{C}} \mathcal{D} (F c, G c)$, where $F$ and $G$ are functors $\mathcal{C} \to \mathcal{D}$.
If you unfold the definition you will find that ...
- 88.7k
8
votes
Accepted
Is taking (co)limits exact in an Abelian category?
In complete generality you only preserve one side of the short exact sequence. Taking colimits is a right exact operation while taking limits is a left exact operation and in general neither is two-...
- 1,549
7
votes
Category Theory: homset preserves limits
There is a rather silk proof which requires some observations:
Let $F$ be a diagram in $\mathcal{C}$ indexed by $\mathscr{J}$. Then, a limiting cone $(\lim F, \mu\colon \Delta \lim F \to F)$ ...
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