(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Consider [tex] f: R^{m+1} - {0} -> R^{(m+1)(m+2)/2}, (x^{0},...,x^{m}) -> (x^{i} x^{j}) [/tex] i<j in lexicografical order

a) prove that f is an immersion

b) prove that f(a) = f(b) if and only if b=±a, so that f restricted to Sm factors through an injective map g from Pm.

c) show g is an embedding

3. The attempt at a solution

for a), [tex] (x^{i} x^{j}) [/tex] is just a one-dimensional image? I think this because it is just the product of two specific components of [tex] (x^{1},...,x^{m}) [/tex], and hence just one-dimensional. But then why does the question state f is sent to some high power of R?

Continuing, under the assumption I was right, something is an immersion if its derivative matrix is injective and it's a map between manifolds. In this case the derivative matrix is just a (m+1)x1 vector, because the image is one-dimensional. Most components are zero, except the derivatives of x^i x^j with respect to x^i and x^j. Because Df(a)=Df(b) if and only if a=b, this would be an injective map. Can anyone check this and tell me if it's half right, I don't feel very confident about it?

for b), I'd say this map is not injective at all since we just kill all components except x^i and x^j, so we could change any component except the ith or jth and still be the same f(a). I must have understood something wrong.

c) was easy. A theorem states that a closed injective immersion is an embedding. because any injective immersion from a compact domain is closed (as in the case with g and Pm), g is an embedding. I also proved Pm was compact, because it is the continuous image of Sm, which is very compact in Rn.

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# Proof that a specific map is an injective immersion

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