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Homotopy theory, homological algebra, algebraic treatments of manifolds.

43 votes
Accepted

Fundamental groups of topological groups.

Here is an example: a product of infinitely many $\mathbb{RP}^\infty$'s. The crucial thing thing to see is that $\mathbb{RP}^\infty$ (or, easier to see, its universal cover $S^\infty$) has a group s …
Todd Trimble's user avatar
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22 votes

to what extent does the category Cov(X) determine a topological space X?

If $X$ is a sober space, you can retrieve $X$ up to homeomorphism from $Cov(X)$. (Nitpick: this is not very good notation; it is very easy to misread it as the category of covering spaces over $X$. I …
Todd Trimble's user avatar
  • 53.3k
21 votes
Accepted

What's special about the Simplex category?

Intuitively, I see the product-preservation or indeed finite limit preservation of geometric realization $\hat{R}: [\Delta^{op}, \mathbf{Set}] \to \mathbf{kSpace}$ as lifting (through the forgetful fu …
Todd Trimble's user avatar
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14 votes

Ends of topological spaces. Why independent of choice of ascending sequence of compact subsets?

Neil has already given adequate reply; this answer is partly for Simon, and partly for those who do like category theory, and realize that its purpose is to make life simpler, not more complicated! …
Todd Trimble's user avatar
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14 votes

Comparisons of convenient categories for algebraic topology

From the nLab (although I was the author of these words): A reasonably large class of examples, including the examples of compactly generated spaces and sequential spaces, is given in the article by …
Todd Trimble's user avatar
  • 53.3k
13 votes
3 answers
704 views

Can a homotopy inverse of the map from a Lie group to loops on its classifying space be give...

Let $G$ be the compact Lie group $SO(n)$. There are some classical constructions of the classifying bundle of $G$ based upon on direct limits of Grassmann and Stiefel manifolds: $$BG \simeq \underse …
Todd Trimble's user avatar
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11 votes
Accepted

is there a universal property that characterises the join of two categories?

It's a special case of what's called a collage or cograph construction. Recall that a profunctor or bimodule between categories $B$, $A$ is a functor $R: A^{op} \times B \to Set$. The cograph of $R$ i …
Todd Trimble's user avatar
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11 votes
Accepted

Simplicial set construction of the classifying space

I believe that's called the Milgram bar construction: R.J. Milgram, The bar construction and abelian $H$-spaces, Illinois J. Math. 11 (1967), 242-250.
Todd Trimble's user avatar
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10 votes

Abelian groups as fundamental groups of topological groups

I'd like to add a quick little explanation to Georges's already sufficient answer (I'm sure this explanation is in Baez's post, but it can be said in a few lines here). One of the morals of the famo …
Todd Trimble's user avatar
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10 votes

A canonical and categorical construction for geometric realization

As to "why is the unit interval the canonical interval?", there is an interesting universal property of the unit interval given in some observations of Freyd posted at the categories list, characteriz …
Todd Trimble's user avatar
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9 votes

Cobordism categories that don't involve manifolds

Here is one type of example, basically inspired by the manifold-type examples but so general that they are not actually categories of manifolds. Let $B$ be any small category with finite coproducts, a …
Todd Trimble's user avatar
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7 votes
Accepted

Reference for Stasheff Operad

Why not look at Stasheff's original paper? He does give a point-set model (where $K_{n+2}$ is a compact convex semialgebraic subset of $\mathbb{R}^n$) and describes explicitly the substitution maps $\ …
Todd Trimble's user avatar
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6 votes

Has this kind of question in topology a special name?

Perhaps the mapping class group of $X$? There is an extensive theory for mapping class groups and their computations. The mapping class group of the (2-dimensional) torus is $SL_2(\mathbb{Z})$.
Todd Trimble's user avatar
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5 votes
Accepted

A $G$-space as a coend

It doesn't seem to be true. Suppose we take $G = \{-1, 1\}$ with the discrete topology, acting on $X = \mathbb{R}$ by usual multiplication. Here $X^G$ consists of a single point $0$. The coend amounts …
Todd Trimble's user avatar
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4 votes

Automorphisms of the rooted tree operad

I think the answer to the question as literally stated is "the trivial group", but I think there are related inquiries which get into some deep combinatorics. One way of thinking about the rooted tr …
Todd Trimble's user avatar
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