Consider an elliptic surface $f :X \rightarrow C$ with $\chi(X) > 0$ or equivalently the fibration has a reduced singular fiber apart from possibly multiple fibers (the field under consideration is $\mathbb{C}$,$X$,$C$ are smooth and projective ). Denote by $\Omega$ the cotangent bundle of $X$ and $K$ the canonical bundle of $C$. We have an inclusion \begin{equation} 0 \rightarrow f^*K \rightarrow \Omega \end{equation} Let $q(X)$ denote the dimension of $H^0(X,\Omega)$ over $\mathbb{C}$. We have $q(X) = g$ where $g$ is the genus of the curve $C$.On the other hand dimension of $H^{0}(C,K)$ is $g$ as well and hence we get (using the above inclusion of sheaves) \begin{equation} H^{0}(C,K) = H^{0}(X,\Omega). \end{equation} Now what I am trying to understand is the above equality in a better way (rather than just by using the dimension argument as above). Is there a way to see why all the 1forms are pull backs of those from the base curve?.

This is a comment starting from a slightly more general context. Most of the following material can be found in a paper of T. Saito and me (but most dealing with the positive characterisitc case). Let $f : X\to C$ be a flat morphism of smooth (geometrically connected) projective varieties over a field $k$ of characteristic $0$. Consider the canonical exact sequence
$$ 0 \to f^*\Omega_{C/k}\to \Omega_{X/k}\to \Omega_{X/C} \to 0. $$
We have $f_{*}\mathcal O_X=\mathcal O_C$. Taking $f_*$ we get an exact sequence of sheaves on $C$:
$$ 0 \to \Omega_{C/k}\to f_{*}\Omega_{X/k} \to f_{*}\Omega_{X/C}\stackrel{\theta}{\to} R^1f_{*}(\mathcal O_X)\otimes\Omega_{C/k}. $$
We have $R^1f_{*}(\mathcal O_X)\simeq \omega_{X/C}^{\vee}$ (in characteristic $0$).
Let $T=\Omega_{X/C, \rm{tors}}$ (torsion as $\mathcal O_X$module).
A local analysis shows easily that $T$ is an invertible sheaf over the verticla divisor $D:=\sum_{s\in C} D_s$, where $D_s=X_s(X_s)_{\mathrm{red}}$ (here again we need $k$ of characteristic $0$) and we have an exact sequence
$$ 0 \to T\to \Omega_{X/C}\to \omega_{X/C}(D) \to S \to 0$$ At the generic fiber $\theta$ is the KodairaSpencer map. It is nontrivial when $f$ is nonisotrivial, and it is injective if moreover the generic fiber has genus $1$. So under these hypothesis, we have
$$ 0 \to \Omega_{C/k}\to f_{*}\Omega_{X/k} \to f_{*}T\to 0. $$
Therefore the canonical map $H^0(C, \Omega_{C/k})\to H^0(X, \Omega_{X/k})$ is an isomorphism if 


Consider the short exact sequence $$ 0\to f^*\Omega_C\to \Omega_X\to \Omega_{X/C}\to 0, $$ and the long exact sequence induced by $f_*$: $$ 0\to \Omega_C\to f_*\Omega_X\to f_*\Omega_{X/C}\to \Omega_C\otimes R^1 f_* \mathscr O_X\to \ldots $$ If $f$ is not isotrivial, then $\beta: f_*\Omega_{X/C}\to \Omega_C\otimes R^1 f_* \mathscr O_X$ is generically injective (at least where $f$ is smooth $f_*\Omega_{X/C}$ is a line bundle and $\beta$ is nonzero). If $\Omega_{X/C}$ is torsionfree, then so is $f_*\Omega_{X/C}$ and hence $\beta$ is injective everywhere. That implies, that then $\alpha:\Omega_C\to f_*\Omega_{X}$ is an isomorphism. Therefore, $H^0(U,\omega_C)=H^0(f^{1}U,\Omega_X)$ for every $U\subseteq C$ open. 

