Return to Answer

Texified because it was on the front page anyway.

Source Link

edited Jun 27, 2014 at 0:10

Kind of late to the party, but the (weak) contractibility follows from pi_i(S^\infty) = 0$\pi_i(S^\infty) = 0$ for i>0$i>0$.

Kind of late to the party, but the (weak) contractibility follows from pi_i(S^\infty) = 0 for i>0.

Kind of late to the party, but the (weak) contractibility follows from $\pi_i(S^\infty) = 0$ for $i>0$.

Source Link

created Oct 11, 2009 at 22:54

Kind of late to the party, but the (weak) contractibility follows from pi_i(S^\infty) = 0 for i>0.