Let $f:\mathbb{ RP}\to\mathbb{ RP}^2$ such that $f([x_1:x_2])=[x_1^2:x_1x_2:x_2^2]$; I have to show that the function is $C^\infty$ and find the differential in any point.
I decided to use these charts for $\mathbb{ RP}$: $$\varphi_1:U_1\to \mathbb R , \ \ \varphi_1([x_1:x_2])=\frac {x_2} {x_1}$$ $$\varphi_2:U_2\to \mathbb R , \ \ \varphi_2([x_1:x_2])=\frac {x_1} {x_2}$$ where $U_1=\mathbb{ RP}\backslash\{[0:1]\}$ and $U_2=\mathbb{ RP}\backslash\{[1:0]\}$. I used these charts for $\mathbb{ RP}^2$: $$ \phi_1:W_1\to\mathbb{ RP}^2, \ \ \phi_1([y_1:y_2:y_3])=\left( \frac{y_2} {y_1},\frac{y_3} {y_1}\right)$$ where $W_1=\mathbb{ RP}^2\backslash\{[0:a:b]\}$ with $a,b\neq 0$, and similarly I defined $\phi_2$ and $\phi_3$.
To show that is $C^\infty$ we can calculate $\phi_1\circ f \circ \varphi_1^{-1} $ and $\phi_3\circ f \circ \varphi_2^{-1}$, since $f(\mathbb{ RP})\subset U_1\cup U_3$; I obtained $\phi_1\circ f \circ \varphi_1^{-1}(s) =\phi_3\circ f \circ \varphi_2^{-1}(s)=\left(s^2,s\right)$, that is clearly $C^\infty$ in every component. However I don't understand what does it mean to find the differential in the coordinates of $\mathbb{ RP}^2$; can you give me an explanation? Thank you in advance
Hint, not an answer
The differential at any point $p \in \Bbb R \Bbb P$ is a linear map from $T_p {\Bbb RP}$ to $T_{f(p)} {\Bbb RP}^2$. And you just know that somehow it's going to involve taking a derivative of $(s^2, s)$ with respect to something, probably $s$, but the exact details are a little vague at present.
But as a first step, you need to decide how you're going to describe elements of each of those tangent spaces. If you don't have a way to do that, then there's really no way to express a function between them.