# Show that a differentiable function is an embedding

Show that $F:\mathbb{P}_\mathbb{R}^{1}\times\mathbb{P}_\mathbb{R}^{1}\longrightarrow\mathbb{P}_\mathbb{R}^{3}$ defined by $F([x_0:x_1],[y_0:y_1])=[x_0y_0:x_0y_1:x_1y_0:x_1y_1]$ is an embedding.

I know that a function is an embedding if $d_pF$ is injective and if $F:M\rightarrow F(M)$ is a homeomorphism.

The problem is that I don't know how to handle the space of derivations in a projective space, and I really have no idea how to start.

• To show the differential is injective, you can check it on each chart. Do you know how to cover $\mathbb{P}^n$ by $n+1$ copies of $\mathbb{R}^n$? Also, are you able to show it's a homeomorphism onto its image? – arkeet Sep 28 '16 at 16:46
• BTW this thing has a name - it is a case of the Segre embedding $\mathbb{P}^{m-1} \times \mathbb{P}^{n-1} \to \mathbb{P}^{mn-1}$. – arkeet Sep 28 '16 at 16:50
• @arkeet I don't know how to cover $\mathbb{P}^n$, and I neither do know how to prove it's a homeomorphism... also I wouldn't know how to effectively show it's injective in each chart since I have never seen any example of how to do it. – Verónica Segura Draper Sep 28 '16 at 16:50
• By cover I just mean it's an atlas. How do you show $\mathbb{P}^n$ is a manifold? – arkeet Sep 28 '16 at 16:52
• @arkeet yes I do know the atlas, for a given $n$, thus I also know how to prove that $\mathbb{P}^n$ is a differential manifold. That is I would take the $U_i=\{[x_0:\dots:x:_n]:x_i\neq 0\}$ and so on. Now how do I proceed? – Verónica Segura Draper Sep 28 '16 at 16:59

In general, a chart $\varphi \colon U \to \mathbb{R}^m$ of a manifold $M$ gives you an isomorphism $T_p M \to T_{\varphi(p)} \mathbb{R}^n \cong \mathbb{R}^n$ for each $p \in U$. So if you also have a chart $\psi \colon V \to \mathbb{R}^n$ of another manifold $N$, if $F \colon M \to N$ and $F(p) \in N$, you can compute $d_p F$ via these charts as a linear map from $\mathbb{R}^m \cong T_p M$ to $\mathbb{R}^n \cong T_{F(p)} N$. Then $d_p F$ is injective iff your linear map $\mathbb{R}^m \to \mathbb{R}^n$ is injective.
So let's say $\varphi_i \colon \mathbb{R}^1 \stackrel\sim\to U_i \subset \mathbb{P}^1$ is the standard atlas of $\mathbb{P}^1$, and $\psi_i \colon \mathbb{R}^3 \stackrel\sim\to V_i \subset \mathbb{P}^3$ is the standard atlas of $\mathbb{P}^3.$ For each chart $U_i \times U_j \subset \mathbb{P}^1 \times \mathbb{P}^1$, you can show that $F(U_i \times U_j) \subset V_{2i+j}$. (For instance, if $p = ([1:x],[1:y]) \in U_0 \times U_0$, then $f(p) = [1:y:x:xy] \in V_0$. So for every $i,j$ you get maps $$\mathbb{R}^2 \xrightarrow{(\varphi_i \times \varphi_j)^{-1}} U_i \times U_j \xrightarrow{F} V_{2i+j} \xrightarrow{\psi_{2i+j}} \mathbb{R}^3,$$ which for instance is $(x,y) \mapsto (y,x,xy)$ when $i = j = 0$. You should be able to show that the differential of this map is injective at every point, for each $i$ and $j$.
Finally, the simplest way I know to show $F$ is a homeomorphism onto its image is this fact: any bijection from a compact space to a Hausdorff space is a homeomorphism. (And here $F$ is a map from a compact manifold to another manifold, which is necessarily Hausdorff.)
• (Proof of the fact at the end: To show $F^{-1}$ is continuous, it's enough to show that $F$ maps closed sets to closed sets. If $A \subset X$ is closed, then $A$ is compact, so $F(A) \subset Y$ is compact. But compact subspaces of Hausdorff spaces are closed.) – arkeet Sep 28 '16 at 19:32