I Natural direction of pushforwards and pullbacks

[ergospherical]

Given a diffeo $\phi : M \rightarrow M'$ (and with $f$ a function on $M'$), vectors $X$ can be "naturally" pushed forward with $\phi_*$ from $T_{p}M$ to $T_{\phi(p)}M'$ subject to $\phi_{*}X(f) \bigg{|}_{\phi(p)} = X(\phi^* f) \bigg{|}_{p}$. And 1-forms $\omega$ are naturally pulled back from $T^*_{\phi(p)}M'$ to $T^*_p M$ subject to $\langle \phi^* \omega, X \rangle \bigg{|}_{p} = \langle \omega, \phi_* X \rangle \bigg{|}_{\phi(p)}$.

Making use of the inverse $\phi^{-1}: M' \rightarrow M$, I think it's possible to also push forward 1-forms ($\omega \mapsto \phi_* \omega$) subject to e.g. $\langle \phi_* \omega, X \rangle \bigg{|}_{\phi(p)} = \langle \omega, {(\phi^{-1})}_* X \rangle \bigg{|}_p$. And similarly I think we can also pull back vectors ($X \mapsto \phi^* X$) subject to e.g. $\phi^* X(f') \bigg{|}_p = X({(\phi^{-1})}^* f') \bigg{|}_{\phi(p)}$, where $f'$ is a function on $M$ [are these right?].

In any case my question is why do vectors seem to naturally be pushed forward, whilst 1-forms and functions seem too be naturally pulled back... is it simply a matter of definition? Thanks

 

[Orodruin]

A tangent vector is naturally pushed forward since it is the tangent of a curve $\gamma$ in $M$ and $\phi\circ\gamma$ is then a curve in $M’$ whose tangent is the pushforward.

 

[cianfa72]

Intuitively in the general case in which $\phi : M \rightarrow M'$ is not injective we cannot define a pullback of a vector field from $M'$ to $M$ the same way we cannot define a pushforward of a scalar field (function) from $M$ to $M'$.

In the latter case which would be the value of the function at the point P in $M'$ having as inverse image through $\phi^{-1}$ different values of the scalar field (function) defined on multiple points in $M$ ?

 

[martinbn]

In any case my question is why do vectors seem to naturally be pushed forward, whilst 1-forms and functions seem too be naturally pulled back... is it simply a matter of definition? Thanks

I'd say for functions is quite clear, because it is just composing with the map. For dual objects it goes the opposite way. Since vectors evaluate on functions, they are pushed forward. One forms evaluate on vectors, so they are pulled back.

 

[fresh_42]

If you have a bijection, then you are automatically disposing of all directions. But bijection in this case means diffeomorphism, which is quite a strong condition.

Pullbacks and pushforwards are dual operators and their existence can be described by commutative diagrams. For short: one is the Jacobi matrix, the other one is its transpose.

Pushforwards are easier to visualize because we can imagine a vector, but not so much a 1-form. It's
$$
(\varphi_*(v))(f)=v(f \circ \varphi) \text{ versus } (\varphi^*\nu)(x) = \nu(\varphi(x))
$$

Your question is a bit like: What if I start with a smooth function $f^*\, : \,M^*\longrightarrow N^*$ on the dual spaces? But don't demand to work this out. I would get lost in directions. However, it's a legitimate setup.

I tried to sort it out here:
https://www.physicsforums.com/insights/pantheon-derivatives-part-iii/
but it is more about definitions than about the why's.