In a handful of contexts people study Calabi-Yau threefolds formed by taking the fiber product of two rational elliptic surfaces. I can't find any detailed explanation of why such geometries are actually Calabi-Yau, so I think it's just a straightforward computation which I don't fully understand. 


Let $\pi: S \to \mathbb{P}^{1}$ and $\pi' : S' \to \mathbb{P}^{1}$ be two rational elliptic surfaces, and define their fiber product


$$X = S \times\_{\mathbb{P}^{1}} S'.$$


With some mild assumptions on $S$ and $S'$, $X$ should be a Calabi-Yau threefold, and I'm hoping someone can help me complete the proof of this. In other words, I want to see that $\omega\_{X}=0$ or $K\_{X}=0$ (however, note that in general, $X$ will certainly not be smooth). 


I believe one should start by considering the obvious map induced by $\pi$ and $\pi'$


$$f: S \times S' \to \mathbb{P}^{1} \times \mathbb{P}^{1}.$$


We can then write $X$ as the pullback of the diagonal $\Delta \subset \mathbb{P}^{1} \times \mathbb{P}^{1}$,


$$X = f^{\*} \Delta.$$


So we can realize $X$ as a hypersurface in $S \times S'$, so you should then be able to apply the adjunction formula:


$$\omega\_{X} = \omega\_{S \times S'}|\_{X} \otimes \mathcal{N}\_{X/S \times S'},$$


where $\mathcal{N}\_{X/S \times S'}$ is the normal bundle of $X$ in $S \times S'$. However, I'm sort of stuck on how to proceed -- How can one explicitly handle both of the two factors in the above tensor product and show they somehow cancel to give 0? 



Show Fiber Product of Rational Elliptic Surfaces is Calabi-Yau
The diagonal $\Delta $ is linearly equivalent to $\{p\}\times \mathbb{P}^1 +\mathbb{P}^1\times \{p\} $ for any $p$ in $\mathbb{P}^1$. Therefore $X$ is the zero locus in $S\times S'$ of a section of $L:=\pi^\*\mathcal{O}(1) \boxtimes \pi'^\*\mathcal{O}(1) $. On the other hand, standard theory of elliptic surfaces gives 
$\omega \_S=\pi ^\*\mathcal{O}(-1) $ and $\omega \_{S'}=\pi' ^\*\mathcal{O}(-1) $, therefore $\omega \_{S\times S'}= L^{-1}$. Then the adjunction formula gives indeed $\omega\_X\cong \mathcal{O}\_{X}$.


