How to prove that $M_{k-12} \to S_k$ is an isomorphism? The following is from the book Modular Forms by W Stein:

My questions:
1- Why multiplication by a nonzero holomorphic $\Delta$ defines an injective map?
2- How showing that "if $f \in S_k$ then $\dfrac{f}{\Delta}\in M_{k-12}$" implies that map is surjective?
I have an undergraduate background in Algebra/Analysis. Simple detailed explanation would be much appreciated.
 A: $\Delta(\tau)$ is nonzero for $\tau$ in the upper half-plane. So if $(g\Delta)(\tau)=0$ then $g(\tau)\Delta(\tau)=0$ and $g(\tau)=0$. Thus $g\Delta$
is identically zero implies $g=0$.
Oh, and $f=(f/\Delta)\Delta$.
A: For Q1, $\Delta$ is non-vanishing on the upper half plane: indeed, since $\Delta$ is a cusp form of weight $12$, using the valence formula (aka the $k/12$ formula) we get $ord_{\infty}(\Delta) = 1$, which implies $\Delta \neq 0$ on $\mathbb{H}.$  Then injectivity is straightforward: $f \Delta = f' \Delta$ implies $ f = f'$. 
For Q2, that's just the definition of surjectivity: given an element of $S_k$, you have produced an element in $M_{k-12}$ which maps to it.
A: For $U$ a  open subset of $\mathbb C$, let $\mathcal O(U)$ be the ring (and $\mathbb C$-vector space) of holomorphic functions on $U$.  If $f \in \mathcal O(U)$, then $\phi: \mathcal O(U) \rightarrow \mathcal O(U)$, $\phi(g) = gf$ defines a linear transformation.
If $U$ is connected, then a standard result from complex analysis is that $\mathcal O(U)$ is an integral domain (since the zeroes of a nonzero holomorphic function are isolated), which implies that $\phi$ must be injective.  
