I'm very new to this whole sheaf business and very confused, so forgive me if the question is dumb (or not, I have no idea really).
Let $\mathscr{F}$ be a quasi-coherent sheaf on $Y$. For simplicity say $Y$ is an affine variety. Suppose we have a map $f:X\to Y$ (suppose $X$ is also an affine variety), I'm wondering if the inverse image (pullback) sheaf $f^{-1}\mathscr{F}$ is quasi-coherent? And if so what is the relation between their corresponding modules?
Let me tell you about my own attempt:
I know that any quasi-coherent sheaf is isomorphic to $\tilde{M}=M\otimes_{k[Y]}\mathscr{O}_Y$ where $M$ is a $k[Y]$-module. The map $f$ also induces a map $f^*: k[Y]\to k[X]$, meaning $k[X]$ is a $k[Y]$-algebra. By extension of scalars we find a $k[X]$-$k[Y]$ bimodule $M\otimes_{k[Y]}k[X]$. Then $$\tilde{M} = M\otimes_{k[Y]}\mathscr{O}_Y =(M\otimes_{k[Y]}k[X])\otimes_{k[X]}\mathscr{O}_Y$$ The only thing stopping the above sheaf to be sheaf on $X$ is $\mathscr{O}_Y$ not being $\mathscr{O}_X$. So I thought maybe we have (very naively) $$ f^{-1}\tilde{M} = (M\otimes_{k[Y]}k[X])\otimes_{k[X]}\mathscr{O}_X $$ I searched a little bit and seems like this is actually true. So I thought of proving it. It is enough we show it for a basis of topology of $X$, say $D(g)=\{x\in X|g(x)\neq 0\}$ for $g\in k[X]$. Then by definition $$ \Gamma(D(g), f^{-1}\tilde{M})=\varinjlim_{V\supset f(D(g))} \Gamma(V, \tilde{M})=M\otimes_{k[Y]} \varinjlim_{V\supset f(D(g))} \Gamma(V, \mathscr{O}_Y) $$ which is $(M\otimes_{k[Y]}k[X])\otimes_{k[X]} \Gamma(D(g), f^{-1}\mathscr{O}_Y)$. This means $$f^{-1}\tilde{M} = (M\otimes_{k[Y]}k[X])\otimes_{k[X]} f^{-1}\mathscr{O}_Y$$ But do we have $f^{-1}\mathscr{O}_Y=\mathscr{O}_X$? I don't so (I mean for any $f$ this is true? really?)! But if it is, how do we prove it? Also in the above argument is there any flaws?