I'm stuck in this problem:
"Let $i:S^2\rightarrow \mathbb{R}^3$ the inclusion. Prove that the map $i_{*}:TS^2\rightarrow T\mathbb{R}^3$ is an embedding".
I consider $p=(x_1,x_2,x_3)\in S^2$ and $X=(X^1,X^2,X^3)\in TS^2$, so \begin{array}{crcl} i: &S^2 & \longrightarrow & \mathbb{R}^3 \\ & p & \longmapsto & i(p)=p \end{array} and \begin{array}{crcl} i_*: & TS^2 & \longrightarrow & T\mathbb{R}^3 \\ & (p,X) & \longmapsto & \big(i(p),i_{*p}(X)\big)=\big(p, J_i X^T\big) \end{array} where $J_i$ is the Jacobian matrix of $i$. If we compute the Jacobian matrix of $i_*$, we get \begin{equation}J_{i_*}=\left( \begin{array}{cc} \frac{\partial(x_1,x_2,x_3)}{\partial (x_1,x_2,x_3)} & \frac{\partial (x_1,x_2,x_3)}{\partial (X^1,X^2,X^3)} \\ \frac{\partial J_i X^T}{\partial (x_1,x_2,x_3)} & \frac{\partial J_i X^T}{\partial (X^1,X^2,X^3)} \end{array} \right)=\left( \begin{array}{cc} J_i & 0 \\ \frac{\partial J_i X^T}{\partial (x_1,x_2,x_3)} & J_i \end{array} \right) \end{equation} Since $i$ is an embedding, it's an immersione and then $J_i$ has full rank, per columns, $rank(J_i)=\dim(S^2)$. So we have that $J_{i_*}$ has full rank, per columns, $rank(J_{i_*})=\dim(TS^2)$ and then $i_*$ is an immersion.
Now I just need to prove that $i_*$ is a homeomorphism into its image, or even just that $i_*(TS^2)$ is an embedded submanifold.
So the question is: how can I prove that $i_*(TS^2)$ is an embedded submanifold?
Thank you so much