$\newcommand{\H}{\operatorname{H}}\newcommand{\F}{\mathcal{F}}\newcommand{\G}{\mathcal{G}}\newcommand{\O}{\mathcal{O}}\newcommand{\I}{\mathcal{I}}$ Let $X$ be a projective scheme of dimension one over $k$ with irreducible components $Z_1,\ldots,Z_r$. Let $\F$ be a coherent (for simplicity we may assume $\F$ to be invertible) $\mathcal{O}_X$-module.
Is it possible to express the dimensions of $\H^0(X,\F)$ and $\H^1(X,\F)$ over $k$ in terms of the dimensions of the cohomologies of the restrictions of $\F$ to the $Z_i$?
What I tried
I only found that the push forward of a sheaf along a closed immersion is compatible with taking cohomology, i.e. if $i: Z \to X$ is a closed immersion and $\G$ a coherent sheaf on $Z$, then $\H^p(X,i_*\G) = \H^p(Z,\G)$. This yields by plugging in $\G = i^{-1}\F$ the equality $\H^p(X,\F \otimes_{\O_X} \O_X/\I) = \H^p(Z,\F_{\mid Z})$ where $\I$ denotes the ideal sheaf defining $Z$.
Do you know any further information that may be helpful in this situation?
Or put in another disguise:
Is there a connection between $\chi(\F)$ and $\chi(\F_{\mid Z_1}), \ldots, \chi(\F_{\mid Z_r})$?
What I tried
For every irreducible component $Z$ of $X$ with ideal sheaf $\I$ we have the exact sequence $0 \to \I \to \O_X \to \O_X/\I \to 0$ and tensoring with $\F$ (using that $\F$ is invertible and thus flat) provides $0 \to \F \otimes_{\O_X}\I \to \F \to \F \otimes_{\O_X} \O_X/\I \to 0$ and therefore $\chi(\F) = \chi(\F \otimes_{\O_X} \O_X/\I) + \chi(\F \otimes_{\O_X} \I)$. But this does not really help me.
For simplicity, let me assume that $X, Z_i$ are reduced (else, a little more care is needed). Then, you have an exact sequence, $0\to\mathcal{O}_X\to\oplus\mathcal{O}_{Z_i}\to K\to 0$, where $K$ is a skyscraper sheaf whose length $\ell$ can be computed if you knew how the components met. Tensoring with $\mathcal{F}$ and taking cohomologies, you get $\chi(\mathcal{F})=\sum\chi(\mathcal{F}_{|Z_i})-\ell$.