Let $X$ be a compact Riemann surface of genus $g \geq 0$. Let $\gamma \in H_1(X , \mathbb{Z})$ and $\omega_1,...,\omega_g$ a basis for the space of holomorphic forms. The next result is in general true?
$$ \int_{\gamma} \omega_i = 0 \text{ for all } i \in \{1,...,g \} \text{ if and only if } \gamma = 0 \text{ in homology.} $$
It's trivially that the $\textbf{If}$ part is true, but the converse?
My attempt: If $c_1,...,c_{2g}$ are a basis for $H_1(X, \mathbb{Z})$, then $\gamma = m_1 c_1 + ... + m_{2g}c_g$ for some $m_i \in \mathbb{Z}$. Now
$$\int_{\gamma} \omega_i = m_1\int_{c_1} \omega_i + ... + m_{2g} \int_{c_{2g}} \omega_i = 0 \quad \forall i$$
and we consider by taking conjugates
$$\int_{\gamma} \overline{\omega_i} = m_1\int_{c_1} \overline{\omega_i} + ... + m_{2g} \int_{c_{2g}} \overline{\omega_i} = 0 \quad \forall i$$
then the matrix
$$P_{2g\times 2g} = \left(\int_{c_j} \omega_1 ,...,\int_{c_j} \omega_{g},\int_{c_j} \overline{\omega_1} ,...,\int_{c_j} \overline{\omega_{g}} \right)_{1 \leq j\leq 2g} $$
has no null kernel. By Riemann's relations $P$ is non singular, this implies $m_1 = ... =m_{2g} = 0$ and $\gamma = 0$.
Note that there is a well defined Hodge star $\star$ on $X$ (locally given by $\star d z = - i d z, \star d \bar z = i d \bar z)$, consider:
Harmonic $1$-forms: $\mathscr{H}^1(X)=\{\alpha \in \Omega^1(X;\mathbb{C})\vert ~d\alpha=0=d\star \alpha\}$
Holomorphic $1$-forms: $H^{1,0}(X)=\{\alpha \in \Omega^1(X;\mathbb{C})\vert \bar \partial\alpha = 0\}$
Antiholomorphic $1$-forms: $H^{0,1}(X)=\{\alpha \in \Omega^1(X;\mathbb{C})\vert \partial\alpha = 0\}$
By the Hodge-decomposition $\ker (d\vert_{\Omega^1})=\mathscr{H}^1(X) \oplus \mathrm{im } (d\vert_{\Omega^0})$ and in particular $$ \mathscr{H}^1(X) \rightarrow H^1_{dR}(X,\mathbb{C})= \ker (d\vert_{\Omega^1}) /\mathrm{im } (d\vert_{\Omega^0}), \quad \alpha \mapsto [\alpha] $$ is an isomorphism and hence the pairing $$ H_1(X;\mathbb{C})\times \mathscr{H}^1(X) \rightarrow \mathbb{C},\quad (c,\alpha) \mapsto \int_c\alpha $$ is non degenerate. Note that $$ \mathscr{H}^1(X) \xrightarrow{\sim} H^{1,0}(X) \oplus H^{0,1}(X), \quad \alpha \mapsto \frac{\alpha + i \star \alpha}{2} \oplus \frac{\alpha - i \star \alpha}{2} $$ is an isomorphism (simple calculation) and that complex conjugation is an isomorphism $$ C:H^{1,0}(X) \xrightarrow{\sim} H^{0,1}(X). $$ Hence if $c\in H_1(X;\mathbb{C})$ satisfies $\int_c \omega_i = 0$ for $\omega_1,\dots,\omega_g$ a basis of $H^{1,0}(X)$, then automatically $\int_c \alpha = 0$ for all $\alpha \in \mathscr{H}^1(X)$ and non-degeneraracy of the pairing shows that $c = 0$.