Differential geometry; evaluating the differential $df$ of a function $f$ from the sphere to a meridian and the first fundamental form Let $C$ be the meridian $C= \{ (x,y,z) \in \Bbb S^2 | y=0,x\geq 0 \}$. Let $f$ map the sphere $\Bbb S^2$ to $C$ such that $f$ maps every point on the sphere to the unique point on $C$ with the same $z$-coordinate. Now I need to show that $f$ is continuous at all points on $\Bbb S^2$ except the north and south pole. Then I need to show that $I_{f(p)}(df_p(v)) \leq I_p(v)$ for all $p \in S= \Bbb S^2\setminus \{N,S\}  $ and $v\in T_pS$. Here $I$ is the first fundamental form.
I think that I got the continuity part. I showed it as follows: Explicitly, $f(x,y,z) = (\sqrt{1-z^2},0,z)$. For some point $p\in S$ there exists a local parametrization $\textbf{x} :U \to V\cap S$ such that $\textbf{x}(q)=p$ for some $q\in U$ where $U$ and $V$ are open subsets of $\Bbb R^2$ and $\Bbb R^3$ respectively. Then define $\textbf{y}:U\to W \cap C$ via $\textbf{y} = f(\textbf{x})$ for some open set $W \subset \Bbb R^3$. Then we define $f$ to be differentiable at a point $p\in S$ if $\textbf{y}$ is differentiable at $\textbf{x}^{-1}(p)$. Now we see that $f$ when seen as a function from $\Bbb R^3$ to $\Bbb R^3$ is a differentiable function expect at $z = \pm 1$. Therefore $\textbf{y}$ is differentiable due to the differentiability of $\textbf{x}$ and $f$ restricted to $S$. Computing the derivatives gives $\textbf{y}_u   = \nabla f \cdot  \textbf{x}_u  = \left(\frac{-z}{\sqrt{1-z^2}} ,0,1 \right)\cdot \textbf{x}_u $ and $\textbf{y}_v =\nabla f \cdot  \textbf{x}_v $.
The next part I do not fully understand though. I want to calculate $df_p(v)$. Now I think it should be $(df_p)_{ij} = \frac{\partial f_i }{\partial x_j}$ where $f = (f_1,f_2,f_3)$ and $(x_1,x_2,x_3) = (x,y,z)$. Then I get that 
$$df_p = \left(\begin{matrix} 0 &0&0\\0 &0&0  \\ \frac{-z}{\sqrt{1-z^2}}&0&1  \end{matrix}  \right) .$$
However then $I_{f(p)}(df_p(v))  =\frac{z^2}{1-z^2}v_1^2 + v_3^2  $ which is not necessarily smaller than $v_1^2+v_2^2+v_3^2 = I_p(v)$. Here $v=(v_1,v_2,v_3) \in T_p S$. What am I doing wrong? I think my understanding of what exactly $df_{f(p)}(v)$ means is not really great and I might be making a mistake here. Thanks!
EDIT: I am very new to differential geometry so if my question seems to not make sense or things are very weird please feel free to let me know. Thanks!
 A: First, your formula for $f$ should be $f(x,y,z) = (\sqrt{1-z^2},0,z)$. So you can try recomputing.
But I would prefer to work in spherical coordinates $(\phi,\theta)$, where (using the mathematician's notation, not the physicist's)
$$(x,y,z) = (\sin\phi\cos\theta,\sin\phi\sin\theta,\cos\phi).$$
Then $f(\phi,\theta) = (\phi,0)$. You can easily check that if $v = a \dfrac{\partial}{\partial\phi} + b\dfrac{\partial}{\partial \theta}$, then $I(v) = a^2 + b^2\sin^2\phi$. But it's easy to see that $df_{(\phi,\theta)}(v) = a \dfrac{\partial}{\partial\phi}$, and $I(df_p(v)) = a^2 \le a^2+b^2\sin^2\phi = I(v)$.
EDIT: If you insist on parametrizing the sphere as a graph, use $(y,z)$ as local coordinates. Then if $p=(\sqrt{1-y^2-z^2},y,z)$, $df_p$ will send $v=a \dfrac{\partial}{\partial y} + b\dfrac{\partial}{\partial z}$ to $b \dfrac{\partial}{\partial z}$. I leave it to you to check that $$I_p(v) = \frac{(1-z^2)a^2+2yzab+(1-y^2)b^2}{1-y^2-z^2},$$
which is in fact greater than or equal to
$$I_{f(p)}(df_p(v)) = \frac{b^2}{1-z^2}.$$
(Their difference is $\dfrac{(a+byz-az^2)^2}{(1-z^2)(1-y^2-z^2)}$!)
