Concrete example illustrating equivalence of two definitions of pushforward In manifold analysis, there are two definitions of pushforward of vectors on Euclidean spaces. Let $\phi$ be a map from $M$ to $N$, which are subsets of two Euclidean spaces. Then the pushforward of $\phi$, denoted $\phi_*$, can be defined "concretely" as the Jacobian of $\phi$, denoted $D(\phi)$. Given a vector $v$ on $M$, expressed as a column matrix, one can then calculate $\phi_*v$ by doing matrix multiplication $D(\phi)v$.
However, $\phi_*$ can also be defined "abstractly" as follows: given a vector $v$ on $M$, which is seen as a partial differentiation operator on real-valued functions on $M$, $\phi_*$ can be seen as an operator transforming $v$ to $\phi_*v$, which is a partial differentiation operator on real-valued functions on $N$ such that if $f$ is any real-valued function on $N$, then we have $\phi_*v(f) = v(f \circ \phi)$.
Can anybody give a concrete example illustrating the equivalence of these two definitions?
 A: Following the suggestion given, I now give an example. Let $\phi: \mathbb{R}^2 \rightarrow \mathbb{R}^3; \phi(x, y) = (u, v, w)$ such that $u = x + y, v = x - y, w = xy$, $v = [v_1, v_2]^T$ be a vector in $\mathbb{R}^2$ expressed as a column vector, $f: \mathbb{R}^3 \rightarrow \mathbb{R}; f(u, v, w) = uvw$. Then
\begin{eqnarray*}
\phi_* = D(\phi) = \left[\begin{array}{rr}
1 & 1 \\
1 & -1 \\
y  & x
\end{array} \right]
\end{eqnarray*}
And so
\begin{eqnarray*}
\phi_*v & = & \left[\begin{array}{rr}
1 & 1 \\
1 & -1 \\
y  & x
\end{array} \right]\left[\begin{array}{r}
v_1 \\
v_2
\end{array} \right] \\
& = & \left[\begin{array}{c}
v_1 + v_2 \\
v_1 - v_2 \\
v_1y + v_2x
\end{array} \right] \\
\end{eqnarray*}
Then we have
\begin{eqnarray*}
\phi_*v(f) & = & \left((v_1 + v_2)\frac{\partial}{\partial u} + (v_1 - v_2)\frac{\partial}{\partial v} + (v_1y + v_2x)\frac{\partial}{\partial w}\right)(uvw) \\
& = & (v_1 + v_2)vw + (v_1 - v_2)uw + (v_1y + v_2x)uv \\
& = & (v_1 + v_2)(x - y)(xy) + (v_1 - v_2)(x + y)(xy) + (v_1y + v_2x)(x + y)(x - y) \\
& = & v_1(3x^2y - y^3) + v_2(x^3 - 3xy^2)
\end{eqnarray*}
On the other hand, since $f \circ \phi(x, y) = x^3y - xy^3$, we have
\begin{eqnarray*}
v(f \circ \phi) & = & \left(v_1\frac{\partial}{\partial x} + v_2\frac{\partial}{\partial y}\right)(x^3y - xy^3) \\
& = & v_1(3x^2y - y^3) + v_2(x^3 - 3xy^2)
\end{eqnarray*}
We have thus verified that $\phi_*v(f) = v(f \circ \phi)$.
