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Nov
21
awarded  Enlightened
Nov
21
awarded  Nice Answer
Nov
21
revised When does $\mathbf{Top}/X$ embedd fully faithfully into $\mathbf{Top}$?
added 3 characters in body
Nov
21
answered When does $\mathbf{Top}/X$ embedd fully faithfully into $\mathbf{Top}$?
Nov
15
revised Every free abelian group is slender, why?
deleted 55 characters in body
Nov
15
answered Every free abelian group is slender, why?
Nov
15
reviewed Leave Open Every homomorphism from the Baer–Specker group into a slender group factors through ${\bf Z}^n$, why?
Nov
15
reviewed Leave Open Every free abelian group is slender, why?
Nov
4
comment Finding a norm on $ \mathbb{R}^X $ such that the “natural” embedding of a metric space $ X $ in $ \mathbb{R}^X $ becomes an isometry
If the image of the Kuratowski embedding fails to be linearly independent, you should be able to tweak it a bit to fix this. For instance, you could enlarge $X$ to $X\times \mathbb{R}$ with (say) the $L^2$ product metric, and I think that should cause any linear dependences you may have had on $X$ to be violated. The only way I can imagine finding a suitable norm on $\mathbb{R}^X$ is by solving your original problem and then using a nonconstructive linear isomorphism between $\mathbb{R}^X$ and your Banach space--as I said before, defining a norm on all of $\mathbb{R}^X$ is hard.
Nov
4
comment Finding a norm on $ \mathbb{R}^X $ such that the “natural” embedding of a metric space $ X $ in $ \mathbb{R}^X $ becomes an isometry
It seems to me like a far more natural choice of $T$ would be $T(x)(y)=d(x,y)$. If $X$ is bounded, this is an isometry with respect to the sup norm on the space of bounded functions from $X$ to $\mathbb{R}$. In general, if $X$ is infinite, I would not expect there to be any natural norm that is well-defined on all of $\mathbb{R}^X$ (in particular, there does not exist a norm that makes every projection continuous).
Nov
4
answered Does a graded vector space isomorphism between the homology of two loop spaces imply the existence of an algebra isomorphism?
Nov
4
comment Order dimension vs topological dimension of a poset
Do you have any reason to think they are related? They seem wildly different. Note that a totally ordered set (such as $\omega_1$) can have uncountable covering dimension.
Nov
3
answered functions which covers(good covers) manifolds
Nov
1
comment Self-similarity for simple algebraic structures
If you mean to ask more questions than the one question you have at the end (which is only even related to about a third of your post), you should state them explicitly.
Oct
31
comment Relations between characteristic classes of a group and the Stiefel-Whitney/Pontryagin classes
I think you misread the question--as I read it, it is asking about relations between the characteristic classes of the tangent bundle of a manifold and characteristic classes of arbitrary $G$-bundles.
Oct
25
awarded  Nice Answer
Oct
23
reviewed Leave Open Is a group uniquely determined by the sets $\{ab,ba\}$ for each pair of elements a and b?
Oct
23
answered Are there examples of families of objects which are canonically isomorphic, but where diagrams of canonical isomorphisms don't commute?
Oct
22
reviewed Leave Open A metric associated with a continuous surjective map $f:X\to Y$
Oct
22
answered A metric associated with a continuous surjective map $f:X\to Y$