What is the pushforward of a function (not a vector)

Question

If we have two manifolds $M$, $N$ with the map $f:M \to N$, then this induces a map between their tangent spaces $f_*:T_pM \to T_{f(p)} N$. By duality, another map exists $f^* : T^*_{f(p)}N \to T^*_pM$. This gives rise to the following relation: \begin{equation} f^* \omega(V) = \omega(f_* V) \tag{1} \end{equation} for $V \in T_pM$ and $\omega \in T^*_{f(p)}N$.

Now what happens when $\omega$ is a smooth function on $N$, i.e. $\omega \in C^\infty(N)$? Then I think we have $f^* : C^\infty(N) \to C^\infty(M)$. If so, what does $f_*$ maps such that it is consistent with equation $(1)$?

Answer 1

$f^*(g)=g\circ f$ when $g\in C^\infty(N)$. There are no tangent vectors here, since functions are $0^{\text{th}}$ order tensors.

Answer 2

In general there is no such thing as a pushforward map. In good situations, such a map exists, and roughly speaking it corresponds to "integration along the fibre" (also known as the Gysin map or the "wrong way map").

Answer 3

Given a smooth function $\omega$ on $N$, by right compositing with $f$, we get $f^*(\omega)=\omega\circ f:M\rightarrow \mathbb{R}$, a smooth map on M. Taking the derivative of this function at a point $p\in M$, we get $D\omega_{f(p)}\circ f_*$, i.e. \begin{equation} D(f^*(\omega))(p)=D\omega_{f(p)}\circ f_* \end{equation} Hope this might be relevant to your question.