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Aug 16, 2013 at 23:48 comment added Ricardo Andrade A minor nitpick: $C^n(T)$ is not a compact space.
Aug 16, 2013 at 21:04 history edited Jeff Strom CC BY-SA 3.0
acknowledge and discuss shortcoming in answer
Aug 16, 2013 at 19:30 review Low quality posts
Aug 16, 2013 at 19:31
Aug 16, 2013 at 19:16 comment added Omar Antolín-Camarena Why is $i_\ast:\pi_1(K) \to \pi_1(C)$ trivial? If for example, $F$ were a homeomorphism $S^n \to K$, sure. But in general, $i_\ast$ need not be trivial, for example, $f$ could squash $S^n$ to an interval and wrap it a along an essential $1$-dimensional loop in $C$. I guess if $C$ were a regular CW-complex (which it almost surely is), applying your observation to the attaching maps does show that $C$ is a $K(\pi,1)$.
Aug 16, 2013 at 19:12 history answered Jeff Strom CC BY-SA 3.0