Let $f:X \to \mathbb{P}^1$ be a flat projective morphism between projective Noetherian schemes. Let $\mathcal{L}$ be line bundle on $X$ such that $f\_\*\mathcal{L}$ is locally free $\mathcal{O}\_{\mathbb{P}^1}$-module. Then, is $f\_\*(\mathcal{L}^\*) \cong (f\_\*\mathcal{L})^\*$, where by $\mathcal{F}^\*$ denotes the dual of a locally free sheaf $\mathcal{F}$? 



Does push forward commute with taking dual?
Since this didn't already show up in the answers, we do have the following if we additionally assume that $R^i f\_\* L = 0$ for $i > 0$. Indeed, then we have


$$\mathcal{H}om\_{\mathbb{P}^1}(f\_\* L, O\_{\mathbb{P}^1}) \simeq R \mathcal{H}om\_{\mathbb{P}^1}(R f\_\* L, O\_{\mathbb{P}^1}) \simeq Rf\_\* R \mathcal{H}om\_{X}(L, f^! O\_{\mathbb{P}^1}).$$


The last isomorphism is Grothendieck duality.


Here $f^!O\_{\mathbb{P}^1}$ can be viewed as simply $$(\omega\_X \otimes f^\* \omega\_{\mathbb{P}^1}^\*) [\dim X - \dim \mathbb{P}^1] \cong \omega\_X \otimes (f^\* O\_{\mathbb{P}^1}(2))[\dim f]$$ where $[\dim f] = [\dim X - 1]$ is a shift (as a complex) by the dimension of the fibers of $f$. Since obviously $L$ is locally free, there are no higher Exts, and so here is the isomorphism we do obtain by taking the zeroth cohomology of $Rf\_\* R \mathcal{H}om\_{X}(L, f^! O\_{\mathbb{P}^1})$.


$$\begin{array}{rl}
(f\_\* L)^\* = & \mathcal{H}om\_{\mathbb{P}^1}(f\_\* L, O\_{\mathbb{P}^1}) \\
\cong & R^{\dim f} f\_\* \big(L^\* \otimes \omega\_X \otimes f^\* O\_{\mathbb{P}^1}(2)\big)\\
 \cong & R^{\dim f} f\_\* \big(L^\* \otimes \omega\_X\big) \otimes O\_{\mathbb{P}^1}(2).
\end{array}
$$


Let me know if this doesn't make sense. 


**Edit:** Whoops, I just noticed that David Stapleton just posted the same thing in the comments.


