Matt
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 Apr16 awarded Popular Question Jul2 awarded Curious Jun16 awarded Yearling Jun6 awarded Popular Question Mar5 awarded Yearling Jun2 comment Tangent space of a projective variety is well-defined @ZhenLin I don't see why we can restrict to the case $Y = X \setminus Z(f)$... Jun2 comment Tangent space of a projective variety is well-defined @Hurkyl I must be missing something. Why? Jun2 asked Tangent space of a projective variety is well-defined May24 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? May24 asked Regarding $S^3$ as the set of all quaternions of modulus $1$ May22 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? May22 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? May22 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 $S^1$? May22 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/… May22 asked Why are the Möbius strip and the boundary of a Klein bottle homotopy equivalent to $S^1$ May22 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 May22 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 May22 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 $pf = f$. May22 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 $\tilde{f} \tilde{g} \simeq \mathrm{id}_{\tilde{X}}$? May22 asked $X,Y$ are locally path connected and path connected, with universal covers $\tilde{X}, \tilde{Y}$. If $X \simeq Y$ then $\tilde{X} \simeq \tilde{Y}$