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May 15, 2014 at 16:50 comment added KConrad @KolyaIvankov: В отличие от использования слова Вы, никогда не пишется по-английски You или Your в середине предложения, только you или your. См. english.stackexchange.com/questions/30185/…
May 15, 2014 at 16:16 answer added Hannes Thiel timeline score: 4
Apr 17, 2014 at 14:50 vote accept Kolya Ivankov
Apr 17, 2014 at 13:32 history edited Ulrich Pennig
added c-star-algebras tag
Apr 17, 2014 at 8:02 answer added Ulrich Pennig timeline score: 4
Apr 16, 2014 at 18:56 comment added Kolya Ivankov @Michael Sorry for late answer. This is the notion of homotopy between morphisms that is employed in KK-theory. AFAIK, it has the best overlap with Karoubi-Villamayor categorical approach to homotopy. So, I just employ what is in play.
Apr 10, 2014 at 21:44 comment added Michael @AndrejBauer: Actually, yes I suppose C*-algebras, being metric spaces, are compactly-generated.
Apr 10, 2014 at 21:40 comment added Michael @AndrejBauer: I'm not sure whether C*-algebras are compactly-generated... but in any case, I retract my comment. Indeed, a $*$-homomorphism $\varphi : A \to C([0,1],B)$ is the same thing as a family $(\varphi_t)_{t \in [0,1]}$ of $*$-homomorphisms $A \to B$ such that $t \mapsto \varphi_t(a)$ is continuous for each $a \in A$.
Apr 10, 2014 at 6:24 comment added Andrej Bauer @Michael: are $C^*$-algebras compactly generated? If so, then the standard topology on $C([0,1], B)$ should give the usual notion of homotopy $H$ because maps $A \times [0,1] \to B$ are the same thing as maps $A \to C([0,1], B)$. No?
Apr 9, 2014 at 23:05 comment added Michael I'm not sure your notion of homotopy is the most natural one. I would say that a homotopy of $*$-homomorphisms $A \to B$ is a 1-parameter family $(\varphi_t)_{t \in [0,1]}$ of $*$-homomorphisms such that $t \mapsto \varphi_t(a) : [0,1] \to B$ is norm-continuous for each $a \in A$. This reduces to the usual notion of homotopy for maps $X \to Y$ in the commutative case. Unless I misunderstand you, what you have written would correspond to some "uniform" notion of homotopy in the commutative case.
Apr 6, 2014 at 20:09 answer added Lior Silberman timeline score: 6
Apr 6, 2014 at 18:25 history edited Kolya Ivankov CC BY-SA 3.0
Minor change of title.
Apr 6, 2014 at 18:17 history edited Kolya Ivankov CC BY-SA 3.0
Made the question satisfy the definition of path-connectedness.
Apr 6, 2014 at 17:54 comment added Kolya Ivankov I think Your version of the question is more correct. Thank You!
Apr 6, 2014 at 17:47 comment added user23860 There is an ambiguity in your question. In your question $\epsilon$ is independent of $f$ and $G$. But the locally path connected means for every homomorphism $f$, there is an $\epsilon>0$ such that if $g$ is a homomorphism and $d(f,g)<\epsilon$, then there is a continuous homotopy between $f$ and $g$. If this second version is what you mean, you should edit your question.
Apr 6, 2014 at 16:55 history asked Kolya Ivankov CC BY-SA 3.0