$T$ is torus. $H_1(T,Z)\to \operatorname{Hom}\Omega^1(T),C)$ is injection. Let $T=\frac{V}{\Lambda}$ be a torus where $V$ is a complex vector space of dimension $n$ and $\Lambda$ is a rank $2n$ lattice in $V$. $\Omega^1(T)$ is the space of holomorphic 1-forms of $T$. $dim_C(\Omega^1(T))=n$. 
Poincare duality says $H_1(T)\cong H_1(T)^\star$ where $\star$ indicates duality. However $H_1(T)^\star$ is isomorphic to de Rham group of rank $2n$ over $Z$.(Correct me if I am wrong here.) The $(\Omega^1(T))^\star$ has rank $2n$ over $R$. $H_1(T)^\star$ is full rank lattice in $(\Omega^1(T))^\star$. 
$\textbf{Q:}$ Should $(\Omega^1(T))^\star$ isomorphic to de Rham group as I want to see there is an embedding $H_1(T) \to(\Omega^1(T))^\star$ by integrating closed curves of $H_1(T)$? And what is the reason that $H_1(T,Z)\to (\Omega^1(T))^\star$ is an embedding?(I knew it is Poincaré duality but I need to apply de Rham isomorphism somewhere.) 
 A: Let $T = \mathbb C^n / \Lambda$ be the complex torus, and let $\{ z_1, \dots, z_n \}$ be a complex coordinates for the $\mathbb C^n$. Then $\Omega^1(T)$ is the $\mathbb C$-vector space generated by the one-forms $\{ dz_1, \dots dz_n \}$.
[To see that this is true, observe that every holomorphic 1-form on $T$ can be written as $g_1(z_1, \dots, z_n) dz_1 + \dots g_n(z_1, \dots, z_n) dz_n$ for holomorphic functions $g_1, \dots, g_n$ on $\mathbb C^n$ that are invariant under translations by lattice vectors in $\Lambda$. By Liouville's theorem for several complex variables, the functions $g_1, \dots, g_n$ must be constant.]
Now suppose that $\vec\lambda_1, \dots, \vec\lambda_{2n}$ is a set of vectors in $\mathbb C^n$ that generate the lattice $\Lambda$ over $\mathbb Z$. Each vector $\vec\lambda_i \in \mathbb C^n$ corresponds to a generator $C_i$ of $H_1(T, \mathbb Z)$: to be more specific, this cycle $C_i$ is represented by the straight-line segment in $\mathbb C^n$ joining the zero vector to $\vec\lambda_i$, which descends to a closed cycle in the quotient space $\mathbb C^n / \Lambda$.
Finally, the cycle $C_i$ acts on one-forms in $\Omega^1(T)$ by integration; that is, the cycle $C_i$ corresponds to the linear functional in $(\Omega^1(T))^\star$ that sends each $\omega \in \Omega^1(T)$ to the integral $\int_{C_i} \omega$. In particular, $C_i$ sends the basis vector $dz_j \in \Omega^1(T)$ to $\int_{C_i} dz_j$, and this is equal to the $j$th coordinate of the lattice vector $\lambda_i$.
Since no two $\lambda_i$'s have identical coordinates, this shows that no two  elements of $H_1(T, \mathbb Z)$ act identically on $\Omega^1(T)$. In other words, the map $H_1(T, \mathbb Z) \to (\Omega^1(T))^\star$ is injective.
