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seen Dec 18 at 22:10

Dec
15
comment Is the suspension space contractible?
Consider upvoting or accepting my answer if it's helpful!
Dec
13
revised Inverse image of a functor
edited body
Dec
13
asked Inverse image of a functor
Dec
13
comment Is the suspension space contractible?
Yes, although if all you need is one counterexample, you could just consider what happens when $X$ is $S^0$, which is just two points.
Dec
13
comment Is the suspension space contractible?
Yes, that looks right. See, for example, "suspension" in Wikipedia. It's a way to start with one topological space and get another one.
Dec
13
answered Is the suspension space contractible?
Dec
13
comment Understanding combinatorics
Repeating the problem is not an explanation! @Jorge, my guess is that the table is round, and cyclic permutations are considered the same; the information about the $n = 3$ case is intended to clarify this.
Nov
30
comment Property of constant function
The cocountable example does not seem to work. If $X$ is $\mathbb{N}$ with the cocountable topology, and $f: \mathbb{N} \to \mathbb{R}$, $n \mapsto n \operatorname{mod} 2$, then $f$ is continuous but not constant.
Nov
29
comment Tensor products in general topology?
I see, so we are looking for a map which is continuous in each variable separately, and which is initial for this condition. This would be analogous to the tensor product of (say) abelian groups, which is an initial bilinear map from the product group.
Nov
29
comment Tensor products in general topology?
I'm not sure what you mean by $f$ and $\mu$ being "bicontinuous". Usually it seems to mean a homeomorphism, which would make the existence of $\phi$ trivial.
Nov
29
comment Question involving Sylow's Thm for $|G|=245$
I think Sylow's Theorem Application. Prove G is Abelian has most of what you want.
Nov
29
comment Question involving Sylow's Thm for $|G|=245$
I'm having trouble following you. The groups you mention are not "abelian since they are cyclic"; they are obviously abelian because they are the direct products of cyclic groups (which are themselves abelian).
Nov
21
comment name this “hybrid” categorical construction
There is, but it's a little complicated. :-) I'll try to find a more interesting example.
Nov
20
comment name this “hybrid” categorical construction
Thank you, this pullback might be what I need. However, it's not clear to me how to define $A^*$ using a pullback: a pullback of the $G_i$ loses the information in the $F_i$, and a pullback of the $G_i F_i$ is discrete. It seems likely that a combination of a pullback and something else will give me the original construction, but we don't have that yet.
Nov
18
asked name this “hybrid” categorical construction
Nov
14
comment locally self-similar topologies
Great, @IncnisMrsi, tell me what you find.
Nov
2
accepted a property of global elements in the category of sets
Nov
2
asked a property of global elements in the category of sets
Oct
18
answered Defining algebras over noncommutative rings
Aug
11
answered Boundary connected sum of manifolds