Why first homology group of torus is isomorphic (as abelian group) to lattice? Let $T＝\Bbb C/Λ$ be a torus($Λ$ is lattice).
I heard first homology group $H_1(\Bbb C/Λ,\Bbb Z)$ is isom as abelian group to $Λ$ via the map
$r→\oint\mathrm dz$ (we integrate across the cycle $r$).
Could you tell me the proof of the fact that 'this map is well-defined and surjective and injective'?
(I could only understand this map is group hom )
 A: First, some generalities. Suppose that $X, Y$ are Riemann surfaces and $f: X\to Y$ is a holomorphic covering map, $\alpha, \beta$ are holomorphic 1-forms on $X, Y$ respectively such that $f^*\beta=\alpha$. Suppose that $b$ is a piecewise-smooth loop in $Y$, more precisely, $b: [0,1]\to Y$ such that $b(0)=b(1)=y$. Fix a point $x\in X$ such that $f(x)=y$. Since $f$ is a covering map, there is a unique path $a: [0,1]\to X$ such that $a(0)=x$ and $f\circ a=b$. Then
$$
\int_a \alpha = \int_b \beta. 
$$
(this is just the formula of change of variables in integrals once you unravel what's written).
Now, suppose that $X={\mathbb C}$, $\Lambda$ is a lattice in  ${\mathbb C}$; I'll regard $\Lambda$ as a group of biholomorphic transformations acting on ${\mathbb C}$ properly discontinuously and freely, so $Y={\mathbb C}/\Lambda$ is a nonsingular complex elliptic curve and the quotient map $f: X\to Y$ is a holomorphic (universal) covering map. The holomorphic 1-form $\alpha=dz$ is invariant under $\Lambda$, hence, it descends to a holomorphic 1-form $\beta$ on $Y$ such that $f^*\beta=\alpha$. I'll fix the base-point $x=0\in X={\mathbb C}$ and let $y=f(x)\in Y$. Take a loop $b$ in $Y$ based at $y$, i.e. $b(0)=b(1)=y$. As noted above, the loop $b$ lifts to a unique path $a: [0,1]\to {\mathbb C}$ such that $a(0)=0$; hence, $a(1)=\lambda\in \Lambda$. We have
$$
\int_b \beta= \int_a \alpha = \int_0^\lambda dz= \lambda
$$
(since $X$ is simply-connected, the integral is independent of a path connecting $0$ and $\lambda$). If loops $b_1, b_2$ in $Y$ are homotopic, then $\int_{b_1}\beta= \int_{b_2}\beta$ (since the form $\beta$ is closed). Thus, we get a well-defined map
$$
\phi: \pi_1(Y,y)\to \Lambda, [b]\mapsto \int_b \beta.  
$$
Since the correspondence between the relative homotopy classes $[b]\in \pi_1(Y,y)$ and the group of covering transformations
$\Lambda$ given by $[b]\mapsto a(1)$ is a group isomorphism, we see that the map $\phi$ is an isomorphism as well. Lastly, the group $\pi_1(Y)$ is abelian, so is naturally isomorphic to $H_1(Y)$.
