6 votes

Is the term "category" in Category theory entirely different from the category in topological spaces?

Perhaps disappointingly, there is no connection between the two uses at all. (Given this, I prefer to use "meager" for "first category," and my impression is that this is pretty ...
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4 votes
Accepted

How do natural transformations capture a "lack of arbitrarity"/uniqueness?

$\newcommand{\hom}{\mathsf{Hom}}\newcommand{\fd}{\mathsf{FDVect_k}}$I’m not sure if it is the “main motivation” in modern terms, but as a historical note I believe Eilenberg and McLane first felt the ...
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  • 8,561
4 votes

What is the right adjoint to the functor $\sf{Psh}\to\sf{Set}$ which evaluates the presheaf on the whole space?

As you have already figured out, the functor $Γ$ is the “global sections” functor that evaluates a presheaf $F$ on the entire space $\mathcal{T}$. The desired right adjoint $∇$ of $Γ$ needs to assign ...
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4 votes

What is the right adjoint to the functor $\sf{Psh}\to\sf{Set}$ which evaluates the presheaf on the whole space?

Here is a method to find the answer without "having to guess". The essential point is that the functor of sections over an open subset $U \subseteq X$ (where $X$ is a topological space) is ...
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4 votes
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What are benefits of fixing a Grothendieck universe?

For example, the category of all small categories (or sets, or groups, etc.) is not small. But the category of $\mathbb{U}$-small categories (or sets, or groups, etc.) is small: in particular, it is $\...
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3 votes

Build a "rich" first-order logic within a given category

I think you'll find it difficult to whip up (or find in the literature) a system which fully has the property you want. The reason is that any such framework would have to (be extremely limited or) ...
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3 votes
Accepted

In a regular category a pullback is a pushout if its sides are regular epi

Looks good to me! You can find a similar argument (albeit a much terser one) in an old answer here. It seems like your main worry is with the claim that the kernel pair $r_0, r_1$ of $f : a \to b$ ...
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3 votes
Accepted

Difference between products and coproducts.

While it is true that given $\{G_i\}_{i \in I}$ you have inclusion maps $G_i \to \prod_{i \in I} G_i$, the product $\prod_{i \in I} G_i$ does not satisfy the universal property of the coproduct if $I$ ...
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3 votes

Difference between products and coproducts.

In addition to Matthias's very nice and explicit proof that the product cannot work as a coproduct, here is another important fact about coproducts for any algebraic structure (abelian groups, groups, ...
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2 votes
Accepted

What is the right adjoint to the functor $\sf{Psh}\to\sf{Set}$ which evaluates the presheaf on the whole space?

Let $\nabla$ be the functor which associates to a set $A$ the following presheaf: $\nabla(A)(X) = A$, and $\nabla(A)(U)$ is a singleton for all proper open subsets $U$, with the restriction maps being ...
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2 votes

What is the right adjoint to the functor $\sf{Psh}\to\sf{Set}$ which evaluates the presheaf on the whole space?

I will denote your topological space by $X$, because otherwise I get confused. A morphism of presheaves $F\to \nabla A$ should be the same as a morphism of sets $FX \to A$. If $\phi$ is a morphism of ...
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  • 1,368
2 votes
Accepted

When representables are adjoints

I think you're overthinking this. Rather than proving that $U$ has a left adjoint, it seems the intention is just to prove it preserves limits. The assumption of completeness is not necessary - it was ...
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1 vote

Proof of Proposition A.2.6.15 in Higher Topos Theory

This question has been answered by Maxime Ramzi in MO. See there for his answer.
1 vote

Why is a closed monidal category enriched over itself?

I figured it out. I consulted the comments, and the following post: A closed symmetric monoidal category is enriched over itself. This answer might be uglier than it could have been. Any new answer is ...
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  • 1,277
1 vote

Canonical morphism $\text{im}(f)\to\text{ker}(g)$ for exact sequence in an abelian category

The kernel of $g$ is defined as the pullback of $g \colon B \rightarrow C$ and the zero morphism $0 \rightarrow C$. Therefore defining a morphism into this kernel amounts to defining two "...
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  • 8,384
1 vote

For two objects X and Y in a category, with Y dominating X, prove that X dominates a Zero Object in our category.

By uniqueness, the composition $0\to X\to 0$ is the identity on $0$, hence $X$ dominates $0$.
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1 vote
Accepted

Cocomplete abelian category with enough injectives has exact coproducts

Let $\mathcal{A}$ be an abelian category with enough injective objects, let $\mathcal{B}$ be the full subcategory of injective objects, and let $[\mathcal{B}, \textbf{Ab}]$ be the category of additive ...
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  • 81.9k
1 vote

Additive category without finite limits

Let $X$ be a manifold and let $\mathcal{V}$ denote the category of (say) smooth vector bundles. This category is additive but not Abelian. In particular, it does not contain kernels of all of its ...
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1 vote
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When the tensor product by a $K$-algebra is a faithful functor on the category of $K$-modules?

Edit I just realized this is not a complete answer, since OP assumed $K$ only to be a commutative ring. My pitfall was the German convention that $K$ denotes Körper (fields), while general rings are ...
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1 vote
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Exercise about exact sequence and pushout

Consider the following commutative diagram: \begin{CD} a@>>>0\\ @VfVV @VVV\\ b @>{g}>> c\\ @VVhV @VVkV\\ b' @>{g'}>> c' \end{CD} By exactness of top row, $g$ is the cokernel ...
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