idea of the star position in pullback, pushforward notation i would like to know if there is some idea behind the position of the star in the pullback, pushforward notation or if it is just some notation without background?
What's the reason for the star to be up at the pullback and not down?*
I've read that there is no real standard notation (Spacetime and Geometry, Sean Carroll) and i wanted to find out what the original idea behind the notation was. I didn't find anything useful, so maybe you know?
Edit: *
 A: Well, I don't really know the reason why the stars are where they are. But here's how I remember it: 


*

*You can push-foward the the tangent space for any morphism of smooth manifolds, and you can pull-back the cotangent space. 

*The cotangent bundle is usually written using the star for dual as $T^*M$. 

*Ergo, pull-backs by smooth maps are $\phi^*$. 


Note that it is a common convention that the "upper" starred objects map things in the "reverse" direction. $\phi: M\to N$ and $\phi^*: TN\to TM$ for smooth manifolds. $A: V\to W$ a linear map between inner-product spaces, the adjoint $A^*:W\to V$ goes the other way. 

If I were to engage in some baseless speculation:
In the context of differentiable manifolds (where this question seems to be motivated), the induced (by a smooth map between manifolds) mapping of the cotangent bundle is the first one that you really need a new notation for. For the tangent bundle, if $\phi: M\to N$ you can just use $d\phi: TM\to TN$. So to me it is quite plausible that someone needed a notation for the induced mapping of the cotangent bundle, and decided to adorn $\phi$ with a star (for reasons unknown). Perhaps there is already an established tradition of adding this type of symbols in the superscript position (perhaps to allow enumerating several maps $\phi^*_1, \phi^*_2, \ldots$). And when it comes time where a symbol for the pushforward is needed, it is actually quite natural to just move the star down as it were.
A: This may be too late but I have the following observation. If $\phi:M\to N$ and $\alpha = \phi^*\beta$, then the transformation law for covectors in coordinates is
$$
\alpha_k = \sum_i \frac{\partial y_i}{\partial x_k}\beta_i ,
$$
where $y=\phi(x)$. Denoting by $D\phi$ the matrix representing the derivative of $\phi$, and thinking of $\alpha$ and $\beta$ as row vectors, the above transformation law in matrix notation reads
$$
\alpha = \beta D\phi .
$$
Clearly, if we think of $\alpha$ and $\beta$ as column vectors, then we have
$$
\alpha = (D\phi)^T\beta .
$$
At least visually, it looks very close to $\alpha=\phi^*\beta$.
