Consider an algebraic elliptic surface $f :X \rightarrow C$ with $\chi(X) > 0$ or equivalently the fibration has reduced singular fibers (the field under consideration is $\mathbb{C}$). Now Hodge theory gives us \begin{equation} q(X) = \dim_{\mathbb{C}}(H^{0}(X; \Omega^1_X)) = \dim_{\mathbb{C}}(H^1(X,\mathcal{O}_X)). \end{equation} On the other hand the $\dim_{\mathbb{C}}(H^1(X,\mathcal{O}_X))$ can be seen to be equal to the genus $g$ of $C$. We also have $\dim_{\mathbb{C}}H^0(C,K_C) = g$ and hence the canonical inclusion map of sheaves \begin{equation} 0 \rightarrow f^{*}K_c \rightarrow \Omega^1_X \implies H^{0}(C,K_c) = H^{0}(X,\Omega^1_X). \end{equation} Now what I seek is to understand this "equality" in a more geometric way. Is there a better way to see (apart from the above method) to see that all the 1 forms in fact come from the base curve?.