Intuition for pullback operation In manifolds and complex geometry there is this thing called the pullback. Usually when I see it, its going backwards on maps that are going forwards. I've been told that it is just a composition of functions. I just need help understanding this concept.
Sometimes an example is better than a thousand words. Let's say you have these forms on $\mathbb{R}^{4}$:
    $$\omega^{1}= y\mbox{d}x-x\mbox{d}y-t\mbox{d}z+z\mbox{d}t,$$
$$\omega^{2}= t\mbox{d}x-z\mbox{d}y+y\mbox{d}z-x\mbox{d}t,$$
$$\omega^{3}= -z\mbox{d}x-t\mbox{d}y+x\mbox{d}z+y\mbox{d}t,$$
$$\omega^{4}= x\mbox{d}x+y\mbox{d}y+z\mbox{d}z+t\mbox{d}t.$$
And you want to pull them back into $S^3$. Then you need an application from $S^3$ into $\mathbb{R}^{4}$. Let's take this one
$$F\left(\psi,\,\theta,\,\phi\right)=\left(\sin\psi\sin\theta\cos\phi,\sin\psi\sin\theta\sin\phi,\sin\psi\cos\theta,\cos\psi\right).$$ This application sends a point of coordinates $\left(\psi,\,\theta,\,\phi\right)$ on the sphere to a point of coordinates $(x,y,z,w)$ on $\mathbb{R}^{4}$ where of course
$$x=\sin\psi\sin\theta\cos\phi,$$
$$y=\sin\psi\sin\theta\sin\phi,$$
$$z=\sin\psi\cos\theta,$$
$$w=\cos\psi.$$
Now to do the pullback you have by definition
$$F^{*}(\omega^{i})=\omega^{i}(F_{*})$$
Then what you have to do is just differentiate 
$$dx = \cos\psi\sin\theta\cos\phi d\psi+\sin\psi\cos\theta\cos\phi d\theta-\sin\psi\sin\theta\sin\phi d\phi$$
$$dy = \cos\psi\sin\theta\sin\phi d\psi+\sin\psi\cos\theta\sin\phi d\theta+\sin\psi\sin\theta\cos\phi d\phi$$
$$dz = \cos\psi\cos\theta d\psi-\sin\psi\sin\theta d\theta$$
$$dw = -\sin\psi d\psi.$$

Then you substitute inside the original formula (substituiting $dx,dy,dz,dw$ along with $x,y,z,w$ of course and you obtain the desired forms pulled back
$$\omega^{1}= \cos\theta\mbox{d}\psi-\sin\psi\cos\psi\sin\theta\mbox{d}\theta-\sin^{2}\psi\sin^{2}\theta\mbox{d}\phi,$$
$$ \omega^{2}= \sin\theta\cos\phi\mbox{d}\psi-\sin\psi\left(\sin\psi\sin\phi-\cos\psi\cos\theta\cos\phi\right)\mbox{d}\theta-\sin\psi\sin\theta\left(\cos\psi\sin\phi+\sin\psi\cos\theta\cos\phi\right)\mbox{d}\phi,$$
$$ \omega^{3}= -\sin\theta\sin\phi\mbox{d}\psi-\sin\psi\left(\sin\psi\cos\phi+\cos\psi\cos\theta\sin\phi\right)\mbox{d}\theta -\sin\psi\sin\theta\left(\cos\psi\cos\phi-\sin\psi\cos\theta\sin\phi\right)\mbox{d}\phi,$$
$$ \omega^{4}= 0.$$
You notice that the last one is 0 because the form was variating only along the radius of the sphere. Anyhow you cannot obtain more than 3 indipendent form on $S^3$ and the first three are indipendent.
I think this example is complicated enough to let you practice, but simple enough to le you grasps what's going on here (which is effectively just the formalization of a change of variables).
The general definition of a pullback: 

Given two morphisms $f:X\to Z$ and $g:Y\to Z$, then a pullback of $f$ and $g$ (if it exist) is two morphisms $p_X:P\to X$ and $p_Y:P\to Y$
$\require{AMScd}$
\begin{CD}
    P @&gt;p_Y&gt;&gt; Y\\
    @V p_X V V= @VV g V\\
    X @&gt;&gt;f&gt; Z
\end{CD}
such that given morphisms $p'_X:P'\to X$ and $p'_Y:P'\to Y$ with 
\begin{CD}
    P' @&gt;p'_Y&gt;&gt; Y\\
    @V p'_X V V= @VV g V\\
    X @&gt;&gt;f&gt; Z
\end{CD}
there exist a unique morphism $\varphi:P'\to P$ such that 
$p'_X=p_X\varphi$ and $p'_Y=p_Y\varphi$.

I suspect that the morphism $\varphi$ sometimes is called pullback, but in category theory it's the two morphisms $p_X:P\to X$ and $p_Y:P\to Y$.

See also: Universal Definition for Pullback