Let $\Lambda\subset\mathbb{C}$ be a lattice, and $f:\mathbb{C}\rightarrow\mathbb{C}/\Lambda$ be the projection. Let $\mathcal{F}$ be the sheaf of holomorphic functions on $\mathbb{C}$.
I want to prove the following: for $z\in \mathbb{C}$, $(f_*\mathcal{F})_{f(z)}\simeq \bigoplus_{\lambda\in \Lambda}\mathcal{F}_{z+\lambda}$, while $\Gamma(f^{-1}(y),\mathcal{F}|_{f^{-1}(z)})\simeq \prod_{\lambda\in \Lambda}\mathcal{F}_{z+\lambda}$, where $\Gamma$ denotes the global section.
I can't get the intuition for why this is true. As far as I can understand, $(f_*\mathcal{F})_{f(z)}=\xrightarrow[f(z)\in U, U\subset\mathbb{C}/\Lambda ]{\lim} \mathcal{F}(f^{-1}(U))$, while $\Gamma(f^{-1}(y),\mathcal{F}|_{f^{-1}(z)})=\xrightarrow[f^{-1}(z)\subset V,V\subset\mathbb{C}]{\lim}\mathcal{F}(V)$ (I believe further sheafification is not needed since the sheaf is supported at points, am I right?),
So when passing to limits, it seems that in the first colimit, the neighborhoods around the points $S :=\{z+\lambda:\lambda\in \Lambda\}$ get shrinked together at the same time, while in the second colimit they can be arbitrarily shrinked around each pointin $S$. Is that the reason why the colimits turn out different?
Can someone give some idea for why it should be true? Any help would be appreciated.