Let $f : E\rightarrow S$ be an elliptic curve over a scheme $S$ with identity section $e : S\rightarrow E$.
Why is it true that $e^*\Omega_{E/S}\cong f_*\Omega_{E/S}$? (I believe these should be naturally/canonically isomorphic, but I don't know why).
Why should this be true intuitively? The first ($e^*\Omega_{E/S}$) is the "restriction" of $\Omega_{E/S}$ to the identity section. Here, I think of the sections of $e^*\Omega_{E/S}$ as sections of the cotangent bundle on $E/S$, "restricted to fibral tangent spaces".
What about $f_*\Omega_{E/S}$? I have a difficult time understanding $f_*\Omega_{E/S}$ geometrically. What is its associated vector bundle? (What is the pushforward of a vector bundle anyway?) I suppose its sections over $U\subset S$ are just holomorphic relative differentials on $f^{-1}(U)$, but this doesn't obviously imply that $e^*\Omega_{E/S}\cong f_*\Omega_{E/S}$.
You have an exact sequence $0\to \Omega_{E/S}(-e(S))\to \Omega_{E/S}\to {\Omega_{E/S}}_{|e(S)}=e^*\Omega_{E/S}\to 0$. Applying $f_*$, you get a long exact sequence, $0\to f_*(\Omega_{E/S})\to e^*(\Omega_{E/S})\to R^1f_*(\Omega_{E/S}(-e(S)))\to R^1f_*(\Omega_{E/S})\to 0$. Both the last two terms are line bundles on $S$ by semicontinuity theorems (and Riemann-Roch) and thus they must be isomorphic. So, we get that the first two terms are isomorphic too.