Push-forward of a smooth function I'm confused about the relation between two concepts: the push-forward of a smooth map between two smooth manifolds and the differential of a smooth real-valued function.
Let $M,N$ be smooth manifolds and $\varphi :M\to N$ smooth. Then the push-forward of $\varphi$ at $p\in M$ is defined as\begin{align*}\varphi _{*p}:T_pM & \to T_{\varphi (p)}N \\
\varphi _{*p}(X_p)(g) & =X_p(g\circ \varphi )
\end{align*}for $g$ any smooth function on $N$.
On the other hand, for a smooth function $f:M\to \mathbb{R}$, i.e. for the special case $N=\mathbb{R}$, the differential of $f$ at $p$ is defined as\begin{align*}d_pf:T_pM & \to \mathbb{R} \\
d_pf(X_p) & =X_p(f).
\end{align*}Starting from these definitions, can we show that $f_{*p}=d_pf$, or are they different objects?
 A: Welcome to MSE!
I think you are right, actually, in some books the push forward is just called differential even if $N$ is not the real line and the push forward function is denoted by $df$. See even wiki https://en.wikipedia.org/wiki/Pushforward_(differential)
If both $M,N$ are euclidean spaces then push forward $df$ is just Jacobian of a function at that point.
Your case is special for $M$ is a manifold and $N$ is real line, and there differential and push-forward is also the same.

That being said, even if I have called push-forward "differential" for a while, to see the connection in this special case is quite refreshing!
A: The tangent space $T_\xi \mathbb R$ is canonically isomorphic to $\mathbb R$ via
$$\alpha_\xi : T_\xi \mathbb R \to \mathbb R,  \alpha_\xi(Z_\xi) = Z_\xi(id), $$
where $id : \mathbb R \to \mathbb R, id(x) = x$. Noting that $T_\xi \mathbb R$ has $\left\{ \frac{\partial}{\partial x} \middle\vert_\xi \right\}$ as a basis, we can alternatively specify $\alpha_\xi$ by
$$\alpha_\xi\left(\frac{\partial}{\partial x} \middle\vert_\xi \right) = 1 .$$
The two maps are related by $\alpha_{f(p)} \circ f_{*p} = d_pf$:
$$(\alpha_{f(p)} \circ f_{*p})(X_p) = \alpha_{f(p)}(f_{*p}(X_p)) = f_{*p}(X_p)(id) = X_p(id \circ f) = X_p(f) = d_pf(X_p) .$$
Update:
$\alpha_\xi$ is nothing else than the differential $d_\xi id$. It is in fact customary to write $d_\xi id = dx_\xi$. The bases $\{ dx_\xi \}$ of $T^*_\xi \mathbb R$ and $\left\{ \frac{\partial}{\partial x} \middle\vert_\xi \right\}$ of $T_p\mathbb R$ are dual in the usual sense. If $d_pf \ne 0$, then $dx_{f(p)}$ is the only isomorphism $\beta : T_{f(p)}\mathbb R \to \mathbb R$ such that $\beta \circ f_{*p} = d_pf$.
More generally, if $g : \mathbb R \to \mathbb R$ is smooth, then one can show that $d_p(g \circ f) = d_{f(p)}g \circ f_{*p}$. This is a sort of chain rule.
A: Short answer: $f_{*p}$ and $d_p f$ (as defined in the question) are different but closely related objects.
As pointed out by Binxu Wang 王彬旭 in one of the comments, $f_{*p}$ maps to $T_{f(p)}\mathbb{R}$, whereas as $d_p f$ maps to $\mathbb{R}$, i.e., they map to isomorphic (as vector spaces) but not identical targets. Formally,
\begin{equation}
f_{*p}(\cdot) = d_pf(\cdot) \left(\frac{\partial}{\partial y}\right)_{f(p)},
\end{equation}
where $\left(\frac{\partial}{\partial y}\right)_{f(p)} \in T_{f(p)}\mathbb{R}$ is the basis vector induced by the (trivial) chart $y = id_N$ on $N = \mathbb{R}$. In other words, $d_pf(X_p) \in \mathbb{R}$ is the component of the vector $f_{*p}(X_p) \in T_{f(p)}\mathbb{R}$ with respect to the basis induced by the chart $y = id_N$.
But of course, by virtue of the isomorphism $c \mapsto c \left(\frac{\partial}{\partial y}\right)_{f(p)}$ between $\mathbb{R}$ and $T_{f(p)}\mathbb{R}$, we can consider $d_pf(X_p)$ and $f_{*p}(X_p)$ the same, up to isomorphism, for any $X_p \in T_pM$.
