Pushforward of a Vector Field on $\mathbb{R}^2$ I am trying to solve the following problem.

$f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ be defined by
  $$f(x,y)=(x^2-y^2,2xy)$$ Compute $f_{*}X$ for
  $X(p)=(4p_{1}+2p_{2},p_{1}+3p_{2})$.

I managed to obtain a result yet I am not sure because I do not think that I know the procedure well. Here is the result I found: $Y=f_{*}X=(20x+14y,10x+11y)$.
Coul you check it?
Now, I found something different by using the fact that $Y=\phi_{*}X=\sum Y^j \frac{\partial}{\partial y_{j}}$ and $Y^{j}$ is given by $\sum_{i=1}^{n} \frac{\partial \phi^j}{\partial x_{i}}X^i$. Using directly this proposition, I calculated that $$Y(x,y)=(8x^2+4y-2xy-6y^2, 8xy+4y^2+2x^2+6xy)$$. Is this correct? I am not convinced personally because I might have done some mistake.   
 A: If $M,\,N$ are smooth manifolds, $f\colon M\to N$ is a smooth map, $V\in\mathfrak{X}(M)$ a vector field, then the push-forward of $V$ need not to be defined: for example if $f$ is not injective, to a point $y\in N$ such that $f^{-1}(y)$ contains more than one point, you would associate more then one vector at $y$. Nevertheless you can always define the push-forward of a vector at a point $x\in M$.
You can verify that your map is not injective, for example $f(1/\sqrt{2},1/\sqrt{2})=f(-1/\sqrt{2},-1/\sqrt{2})=(0,1)$, therefore you cannot define a vector field on $N$ pushed forward by $f$. Instead you can calculate the push-forward of a vector $v\in T_pM$ by $f$: $f_{*,p}v(h)=v(h\circ f)$, where $h\in\mathcal{C}^\infty_N(N)$; in this case you can choose local charts $(U,\phi)$ on $X$ and $(V,\psi)$ on $Y$, then the push-forward $f_{*,p}$ is represented with respect to these charts by the jacobian of $\hat{f}=\psi\circ f\circ \phi^{-1}$ in $p$, and the push-forward of a vector at a point $p$ is just the product of the representing matrices in these charts
\begin{equation}
J_p(\hat{f})v=\begin{bmatrix}
\frac{\partial \hat{f}_1}{\partial x_1}(p) &\dots & \frac{\partial \hat{f}_1}{\partial x_m}(p)\\
\vdots & \ddots &\vdots\\
\frac{\partial \hat{f}_n}{\partial x_1}(p) &\dots &\frac{\partial \hat{f}_n}{\partial x_m}(p)
\end{bmatrix}\begin{bmatrix}
v_1\\
\vdots\\
v_m\end{bmatrix},
\end{equation}
where $\phi(p)=(x_1(p),\dots,x_m(p))$.
In you case, the push-forward of the vector $X_p$ by $f$ at a point $p=(p_1,p_2)\in\mathbb{R}^2$ is
\begin{equation}
\operatorname{J}_p(f)X_p=\begin{bmatrix}
2p_1 &-2p_2\\
2p_2 &2p_1\end{bmatrix}\begin{bmatrix}
4p_1+2p_2\\
p_1+3p_2\end{bmatrix}=\begin{bmatrix}8p_1^2+2p_1p_2-6p_2^2\\
2p_1^2+4p_2^2+14p_1p_2\end{bmatrix}.
\end{equation}

I'd like to get some sort of feedback: does my post answer to your question? If no, why? May I improve it? If yes, why haven't you accepted it yet? Thanks.
