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 Jul2 awarded Curious Nov10 awarded Popular Question Oct13 revised $S^n \backslash S^m$ homotopy equivalent to $S^{n-m-1}$ added 69 characters in body Oct13 revised $S^n \backslash S^m$ homotopy equivalent to $S^{n-m-1}$ added 42 characters in body Oct13 asked $S^n \backslash S^m$ homotopy equivalent to $S^{n-m-1}$ Oct13 asked Homotopy equivalence definition Oct13 asked Which of the letters of the alphabet are contractible? Oct11 comment Basic property of homotopy Great, that makes sense. I'll prove that now, and adopt this notation for the future. Thanks a lot Oct11 comment Basic property of homotopy Is that the identity on $X$ or on $I$? I've never seen this notation before. Thanks Oct11 revised Basic property of homotopy edited body Oct11 asked Basic property of homotopy Oct10 accepted Antipodal map on $S^n$ homotopic to identity map if $n$ is odd Oct7 comment Antipodal map on $S^n$ homotopic to identity map if $n$ is odd Thanks, but I don't understand what you mean by "Notice that, since we had to rely on complex numbers, this does not work for even spheres." Oct7 asked Antipodal map on $S^n$ homotopic to identity map if $n$ is odd Aug26 comment n-ary derived operation in universal algebra Thanks - I'd assumed it was standard notation. Aug26 revised Menger's Theorem (updated) edited title Aug26 revised Menger's Theorem (updated) added 365 characters in body Aug26 comment Menger's Theorem (updated) I think I understand now, actually. We might be tempted to say "suppose all $a$-$b$ separators have size $\geq k$. Find a maximum-sized family of independent paths, and choose a point from each of these paths. Then the set of these points is an $a$-$b$ separator, and so there must be at least $k$ paths in the family". But the part in italics isn't necessarily true. Take, for example, a hexagon with one diagonal and have $a$ and $b$ one vertex clockwise from the start of the diagonal at each end. Aug26 asked Menger's Theorem (updated) Aug26 comment n-ary derived operation in universal algebra What are the $x_i$?