Why is this map from Leray spectral sequence zero? In Stacks Project Lemma 32.37.7 (0C0T), one considers a finite morphism $f:X\rightarrow Y$ of schemes and the Leray spectral sequence for $\mathcal O_X^*$ and $f$. It induces an edge map
$$H^1(X, \mathcal{O}_ X^*) \longrightarrow H^0(Y, R^1f_*\mathcal O_ X^*).$$
Then it claims that Stacks Project Lemma 30.17.1 (0BUT) shows the above edge map is zero.
The cited lemma 30.17.1 says that for any invertible $\mathcal O_X$-module $\mathcal L$ and $y\in Y$, there exists an open neighbourhood $V\subseteq Y$ of $y$ such that $\mathcal L\mid_{f^{-1}(V)}$ is trivial. I cannot perceive how this says that the above edge map is zero. I cannot even tell what the image of an invertible module under the edge map is.

Any help is sincerely appreciated. Thanks in advance.
 A: I believe Tag 0BUT implies something stronger, namely that $R^{1}f_{\ast}\mathcal{O}_{X}^{\ast} = 0$. One can see this as follows: Let $\mathcal{F}$ be the presheaf on $Y$ given by $V \mapsto H^{1}(f^{-1}(V),\mathcal{O}_{X}^{\ast})$. Then $R^{1}f_{\ast}\mathcal{O}_{X}^{\ast}$ is the sheafification of $\mathcal{F}$. By Tag 0BUT, for any open subset $V \subseteq Y$ and $s \in \Gamma(V,\mathcal{F})$, there is an open cover $V = \bigcup_{j \in J} V_{j}$ such that $s|_{V_{j}} = 0$ for each $j \in J$. This means that the universal separated presheaf associated to $\mathcal{F}$ is $0$, so the sheafification is $0$ as well.
A: Edit : I assumed that $\mathcal O^*_X$ is quasi-coherent as $\mathcal O_X$-module which is wrong. However I think the argument should still holds since proper base change and Grothendieck theorem are valid in greater generality (i.e for sheaves of abelian groups).

If you can use proper base change, then for any sheaf $F \in \mathcal O_X \text{-mod}$ you obtain $(R^1f_*F)_y \cong H^1(f^{-1}(y), F_{|f^{-1}(y)})$ where if $i : Z \to X$ is the inclusion of a closed subset then $F_{|Z}$ means $i^* F$. Since $f$ is finite $\dim f^{-1}(y) = 0$ and so by Grothendieck's vanishing (see here) we obtain $H^1(f^{-1}(y), F_{|f^{-1}(y)}) = 0$. 
