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I have a homework Question which asks to show that the map

\begin{equation}f:R^3\rightarrow R^4, f(x,y,z)=(x^2-y^2,xy,xz,yz)\end{equation}

Induces an embedding of $RP^2$ into $R^4$

Overall I have a fairly good idea of how I want to go about showing this, however I am looking for a "neat" way of showing $f$ is injective (take neat to mean whatever you want in this setting).

I have an idea that is probably far reaching, but I was wondering if we could somehow use the kernel of $f$ in our argument? If we had a group homomorphism that would be one thing, but this is not so. However it does seem like the only element that gets mapped to zero is $(0,0,1)$ subject to the domain $S^2$, so maybe that's something? Honestly I'm just really lazy and don't want to brute force algebra onto this function to see that it is injective.

I am just looking for some good hints here, anything is appreciated!

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2 Answers 2

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$f(\mathbf{x})=f(\mathbf{y}) \iff \mathbf{x} \sim \mathbf{y}$, where $\sim$ is the relation induced by $\mathbf{x} \sim \lambda \mathbf{x}$ for all $\lambda \in \mathbb R-\{0\}$ (or they are both zero)

Hence, you can show that it is injective, by the universal property of the quotient topology, which essentially says that it is a "final topology," the coarsest one that makes such a quotient continuous.

Hence, $f$ induces a unique continuous map $\tilde{f}:\mathbb R P^2 \to \mathbb R^4$, and it is injective by the first line.

I'm not sure that this avoids the use of "brute force," but it is easier to check that only elements "on the same line" will evaluate to the same vector under $f$.

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  • $\begingroup$ Isn't a final topology the finest one that makes something continuous rather than the coarsest? $\endgroup$ Commented Mar 4 at 16:02
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See Section 2 Example of (Embedding of $RP^2$ into $R^4$): https://www.cs.uic.edu/~sxie/lecture_notes/diff_manifolds/lecture10.pdf

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