I prefer a somewhat different view of ends and coends, with the intuition stemming more from classical linear algebra and functional analysis. So for me an end is really an integral in a categorical ...

perhaps he is implying some even stronger result He is referring to the following result of Peter Freyd (Freyd uncertainty principle): The homotopy category of spaces $HoTop$ does not admit a faithful ...

Antisymmetrization and symmetrization of tensors. Should we divide it by $(n!)$ ? This affects the relations between tensor and (anti-)symmetric algebra, the theory over $\mathbb Q$ and $\mathbb Z$ ...

A complex polynomial is uniquely determined by its set of roots together with multiplicities. This means that the semigroup of your polynomials is freely generated by the set of point on the unit ...

As requested, making comment into the answer. I think Helemskii's book "Lectures and Exercises on Functional Analysis" contains a very nice intro into category theory. It may be a bit light on the ...

The definition you gave is not of an $A_\infty$-space, but just of $A_1$-space. As Charles noted, these two classes of spaces are very different in general. For example, there are also higher ...

Ok, this discussion has grown beyond the level of comments so I'll collect the facts here. A bit of terminology: a $(-1)$-connected space is a space with a choice of basepoint and the category of $(-1)... View answer Accepted answer 11 votes N.B.: I have reread your question and it occured to me that you a probably asking something entirely different. However since I'm unclear what exactly is your question and since I don't want to delete ... View answer Accepted answer 10 votes As per Qiaochu Yuan's comment we need to only understand the space of based maps between$K(A,n)$with a chosen base point. The loop-deloop pair of functors establish an equivalence between the ... View answer Accepted answer 10 votes Firstly,$\mathcal{O}_X |_U = \mathcal{O}_U$(restriction of sheaves), so you could just consider$U=X$. Then$\mathrm{Aff}(X) := \mathrm{Spec}\,\mathcal{O}_X(X)$is the universal affine scheme with a ... View answer Accepted answer 8 votes I don't believe there are non-trivial examples of this concept. Assume that$e: X \to X$is a constant map, then$e$is idempotent. Any category can be embedded fully faithfully into a Cauchy complete ... View answer 8 votes Here is a sketch of the proof, some details filled below. All categories are$(\infty,1)$-categories and all functors are$(\infty,1)$-functors unless specified otherwise. The notion of a topological ... View answer 8 votes I guess this is not the expected kind of answer, but you can study a metric space as a category, enriched over$\mathbb{R}_+$- a monoidal category of non-negative real numbers (with$\infty$) with ... View answer Accepted answer 8 votes We have an equivalence of categories$Aff\simeq Ring^{op}$and a pair of adjoint functors $$\mathcal{O}:Sch\rightleftharpoons Ring^{op} : Spec$$ $$\mathcal{O} \dashv Spec$$ The category of affine ... View answer 8 votes I like to view Yoneda's lemma as a generalization of the description of Galois coverings in topology. To any functor$F: C \to Set$we can associate its category of elements$El(F)$. Its objects are ... View answer 7 votes As already noted above, any category can be considered concrete after a base change to a suitably large universe. However, doing so would be completely missing the point of concreteness. The ... View answer Accepted answer 7 votes See a paper by G. Segal, "K-homology theory and algebraic K-theory" (I'm sure it's not the original source, though). There is a homotopy equivalence between$BU\times \Bbb Z$and the space$Fred$of ... View answer 7 votes I will consider overcategories of the form$Top/S$,$S\in Top$. Their objects are morphisms$R\to S$, their morphisms are triangles, commutative up to a (specified) homotopy, etc. For details see this.... View answer 6 votes The isomorphism you stated exist not only on the level of topmost exterior power, but also on the level of the whole exterior algebra, considered as a superalgebra. For an exact sequence $$0 \to X \... View answer 6 votes The method of forcing in mathematical logic. If you want to prove the consistency of axiom systems, you can just explicitly present a model of it. To prove results about set theory itself, like the ... View answer Accepted answer 5 votes Yes, there is a certain sense in which your statements are true. As Mike Shulman and Qiaochu Yuan said, the strict fiber of a map cannot be defined in HoTT and doesn't make sense, but you can work ... View answer Accepted answer 5 votes Firstly, note that it is enough to construct an isomorphism$$\mathcal{C}(L(A,B),C) \simeq \mathcal{B}(B, R_2(A,C))$$The third natural isomorphism then follows automatically. Secondly, standard ... View answer Accepted answer 5 votes It is a general fact that if you consider an abelian group$A$in topos$T$, then equivalence classes of$A$-torsors in$T$are classified by cohomology group$H^1(T;A)=Ext^1_T(\mathbb{Z},A)$. Here$\...

Let $X = BG \times [0; 1]$, then $\ast_a = (\ast; 0)$, $\ast_b = (\ast; 1)$, $\gamma$ is the image of $[0;1]$, $f_a$ is any choice of path in the class of $a$ and $f_b$ is any choice of path in class $... View answer 4 votes Ok, I have no idea how to fix the invariant of string diagrams, but I have an explicit counterexample. Consider the 2-category of distributors$Dist$: its objects are small categories, its morphisms ... View answer Accepted answer 4 votes I'm not sure if it is the most elegant way, but it is certainly the most direct. So, we need to define the image of$\mathrm {surf}$as an element of$refl_{\mathrm N} = refl_{\mathrm N}$. We proceed ... View answer 4 votes I don't have any direct reference for the notion that you are describing, however the notions of$E_n$-algebras and (topological) operads are very close. Firstly, you should note that you need ... View answer 4 votes Besides the important examples of topoi already mentioned (like SDG), I would argue that the most important topos for any analyst is just the topos of sheaves on some topological space. I assume that ... View answer 4 votes That's not really much of a problem, but still a nice stumbling block for inexperienced and an example of the evilness of formal symbol manipulations, even innocent-looking ones. Consider a function$...
Geometric realization is a functor, so a functor $C \to C^\prime$ induces a morphism of realizations. Since a natural transformation is a functor $C \times (\bullet \to \bullet) \to C^\prime$, any ...