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visits member for 2 years, 5 months
seen Jun 2 '12 at 16:22

Jul
2
awarded  Curious
Jun
16
awarded  Yearling
Jun
6
awarded  Popular Question
Mar
5
awarded  Yearling
Jun
2
comment Tangent space of a projective variety is well-defined
@ZhenLin I don't see why we can restrict to the case $ Y = X \setminus Z(f)$...
Jun
2
comment Tangent space of a projective variety is well-defined
@Hurkyl I must be missing something. Why?
Jun
2
asked Tangent space of a projective variety is well-defined
May
24
comment Regarding $S^3$ as the set of all quaternions of modulus $1$
@QiaochuYuan Ok, thanks. How would I see that there's a subgroup of the group of homeomorphisms of $S^3$ isomorphic to $Q_8$? I can see an obvious action of $Q_8$ on $S^3$ using this identification, but how do I show the action is continuous?
May
24
asked Regarding $S^3$ as the set of all quaternions of modulus $1$
May
22
comment Why are the Möbius strip and the boundary of a Klein bottle homotopy equivalent to $S^1$
Great, thanks. How does $A \cap B$ deformation retract onto a circle?
May
22
comment Why are the Möbius strip and the boundary of a Klein bottle homotopy equivalent to $S^1$
@JimConant Is there a way to visualise these on the unit square representations?
May
22
comment Why are the Möbius strip and the boundary of a Klein bottle homotopy equivalent to $S^1$
@JimConant Oops, sorry. I mean to ask why does $A \cap B$ deformation retract onto $S^1$?
May
22
comment Given two coprime integers, find multiples of them that differ by 1
Run Euclid's algorithm. Then reverse it. See here: en.wikipedia.org/wiki/…
May
22
asked Why are the Möbius strip and the boundary of a Klein bottle homotopy equivalent to $S^1$
May
22
comment Finding a generator of $(\mathbb Z/p\mathbb{Z})^*$
If $p$ is of the form $2^\alpha + 1$ then the primitive elements are precisely the quadratic non-residues
May
22
comment $X,Y$ are locally path connected and path connected, with universal covers $\tilde{X}, \tilde{Y}$. If $X \simeq Y$ then $\tilde{X} \simeq \tilde{Y}$
@t.b. Yes, that's what I meant. Sorry if I'm being dim, but I'm still confused. Clive and I have both shown that $p \tilde{f} \tilde{g} \simeq p$, but we haven't (I don't think) shown that $p \tilde{f}\tilde{g} = p$. I know there's a unique covering translation for a given pair of points, but I cannot see how $\tilde{f} \tilde{g}$ is a covering translation
May
22
comment $X,Y$ are locally path connected and path connected, with universal covers $\tilde{X}, \tilde{Y}$. If $X \simeq Y$ then $\tilde{X} \simeq \tilde{Y}$
@t.b. Actually, sorry. Why is $\tilde{f} \tilde{g}$ a deck transformation? My definition is that a deck transformation is a function $f$ with $pf = f$.
May
22
comment $X,Y$ are locally path connected and path connected, with universal covers $\tilde{X}, \tilde{Y}$. If $X \simeq Y$ then $\tilde{X} \simeq \tilde{Y}$
@t.b. Great, thanks. So it looks like what I'm doing is generally correct. But I'm unsure about the last step in Clive's answer; how do I go from $p \tilde{f} \tilde{g} \simeq p$ to $\tilde{f} \tilde{g} \simeq \mathrm{id}_{\tilde{X}}$?
May
22
asked $X,Y$ are locally path connected and path connected, with universal covers $\tilde{X}, \tilde{Y}$. If $X \simeq Y$ then $\tilde{X} \simeq \tilde{Y}$
May
22
accepted $f:X \to Y$ is a homotopy equivalence if there exists $g,h : Y \to X$ with $fg$ and $hf$ homotopy equivalences