Category theory problem? Linear Algebra problem? Pull-back transformations I'm having a hard time solving this. I'm taking a proof-based undergraduate linear algebra course that has no assigned textbook and this has been making things a little hard since the only resource I have are my notes from lecture -- he's doing things his own way (starts with set theory, then group theory, et cetera). I don't seem to be able to find anything that gives some insight to this problem. 
Could someone give me a hint on how to solve this problem or point me to some material online that could help? Maybe notes in PDF or something? Oh also, what exactly is meant by "Pullback" Thank you. 
$\mathbb{R}(X)$, $\mathbb{R}(Y)$ is the set of functions that are mapped to $\mathbb{R}$. For example,       $\mathbb{R}(X)=\{ f : X\to\mathbb{R} \}$ and $\mathbb{R}(Y)=\{ g : Y\to\mathbb R \}$ where $\mathbb R$ are the real numbers.
Edit: I have figured out that given a $g$ in $Y$, $f = g \cdot \varphi = \varphi^*(g)$, but I don't see how a,b and thus c are true. 
 A: The strategy of proof is very similar for all of the three points: I'll write down (a), and if you want I can write more also regarding (b) and (c).
So, for (a): say that $\varphi$ is surjective. Then we want to show that $\varphi^*$ is injective, i.e. $\varphi^*(f)$ equals the zero map only if $f$ already was the zero map.
Indeed, suppose $\varphi^*(f)(x)=f\cdot\varphi=0\ \forall x$. Then it must be $f(x)=0\ \forall x$ in the image of $\varphi$. But this image is the whole space! Then $f$ must have been the zero map already - you don't lose any domain by precomposing.
On the other hand, suppose your $\varphi^*$ is injective: then $\varphi^*(f)(x)=f\cdot\varphi=0\ \forall x$ implies $f=0$. But now you could for example argue ad absurdum - pick a point which is not in the image of $\varphi$ and a function which is not zero only in that point - then $\varphi^*(f)=0$, but $f$ was not. Then surjectivity must hold, and the two conditions are equivalent.
You should approach (b) and (c) with the same spirit. I hope this helps!
