I have a couple of questions on a proof from Daniel Huybrechts' Complex Geometry Complex Geometry on pages $134:$
Corollary 3.3.6 If $X$ is a compact Kahler manifold, then $Pic^°(X)$ is in a natural way a complex torus of dimension $b_1(X)$.
Proof. We use the bidegree decomposition (Corollary $3.2.12$): $H^1(X, \mathbb{C}) = H^{1,0}(X) \oplus H^{0,1}(X)$, the fact that $\overline{H^{1,0}(X)} = H^{0,1}(X)$, and $H^1(X, \mathbb{C}) = H^1(X, \mathbb{R}) \otimes_{\mathbb{R}} \mathbb{C} =(H^1(X, \mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{R}) \otimes_{\mathbb{R}} \mathbb{C}$. This shows that $H^1(X, \mathbb{Z}) \to H^1(X, \mathbb{C}) \to H^{0,1}(X)$ is injective with discrete image that generates $H^{0,1}(X)$ as a real vector space. (???) In other words, $H^1(X, \mathbb{Z}) \subset H^{0,1}(X)$ is a lattice. Thus, it suffices to show that this inclusion $H^1(X, \mathbb{Z}) \subset H^{0,1}(X)$ coincides with the one given by the exponential sequence. (???) But this is immediate from Lemma $3.3.1.$
Q1: I don't understand the argument why the composition $H^1(X, \mathbb{Z}) \to H^1(X, \mathbb{C}) \to H^{0,1}(X)$ is injective with discrete image.
The first map $H^1(X, \mathbb{Z}) \to H^1(X, \mathbb{C})$ is induced by canonical inclusion $ \mathbb{Z} \to \mathbb{C}$ of locally constant sheaves. Why ist this map injective? Obviously the image is real and since $\overline{H^{1,0}(X)} = H^{0,1}(X)$ it is fully contained in $H^{1,0}(X)$, that's fine. It's also ok that the image is discrete as $H^1(X, \mathbb{Z})$ is finitely generated as $ \mathbb{Z}$- module . now, why it generates $H^{0,1}(X)$ as $ \mathbb{R}$-space?
Recall, that on previous page $133$ between Lemma $3.3.1$ and Prop $3.3.2$ occurs a similar statement that $H^1(X, \mathbb{R}) \to H^1(X, \mathbb{C})$ is injective. I think that the argument that this map is injective is quite similar to that one that $H^1(X, \mathbb{Z}) \to H^1(X, \mathbb{C})$ from above is injective. One possible argument that came me in mind was to use flatness of $\mathbb{Z} \subset \mathbb{R} \subset \mathbb{C}$. Huybrechts nowhere in his book used this terminology thus I think he uses another argument for injectivity of $H^1(X, \mathbb{Z}) \to H^1(X, \mathbb{C})$. Which one?
Q2: Why does suffices to show that the inclusion $H^1(X, \mathbb{Z}) \subset H^{0,1}(X)$ from Q_1 coincides with that one given by the exponential sequence (page 132) to show the claim?
As you noted, $H^1(X,\mathbb Z) \to H^1(X,\mathbb C)$ is induced by inclusion of constant sheaves $\mathbb Z\to\mathbb C$. If one looks at the short exact sequence of sheaves $$0\to\mathbb Z\to\mathbb C\to\mathbb C/\mathbb Z\to 0$$
Where $\mathbb C/\mathbb Z$ is the quotient sheaf, coinciding with the constant sheaf of quotient group. Then using long exact sequence of cohomology, the injectivity of $H^1(X,\mathbb Z) \to H^1(X,\mathbb C) $ is equivalent to $H^0(X,\mathbb C/\mathbb Z) = \mathbb C/\mathbb Z\to H^1(X,\mathbb Z) $ being the zero map. Which follows from surjectivity of $H^0(X,\mathbb C)=\mathbb C\to \mathbb C/\mathbb Z$.
Now $H^1(X,\mathbb Z) $ generates $H^{0,1}(X)$ as real vector space because $H^1(X,\mathbb R) \cong H^1(X,\mathbb Z) \otimes_{\mathbb Z} \mathbb R$ (and similarly for $\mathbb C$). Let $\alpha\in H^{0,1}(X)$, and $\bar\alpha\in H^{1,0}(X)$, then $\alpha+\bar\alpha\in H^1(X,\mathbb R)$ and is generated by integral cohomology, say $\alpha+\bar\alpha=\sum a_i\omega_i$. Then one can just project $\alpha+\bar\alpha$ back to $H^{0,1}(X)$ and recover an expression for $\alpha$ (since $\bar\alpha$ vanishes under the projection $H^1(X,\mathbb C) \to H^{0,1}(X)$.)
For Q2, recall that we wish to consider $\mathrm{Pic}^0(X)$, which is defined using exponential sheaf sequence as the kernel of first Chern class map $c_1: H^1(X,\mathcal O_X^*) \to H^2(X,\mathbb Z)$. It is the same as the image of $H^{0,1}(X)\cong H^1(X,\mathcal O_X) \to H^1(X,\mathcal O_X^*) $, so by first isomorphism theorem it is isomorphic to $H^1(X,\mathcal O_X) $ quotient by the image of $H^1(X,\mathbb Z) \to H^1(X,\mathcal O_X)$. But this map is the inclusion of lattice as discussed above, so $\mathrm{Pic}^0(X)$ is isomorphic to a complex vector space quotient by lattice, which is a complex torus.
By the way, the complex dimension of the complex torus should be $b_1(X)/2$ instead of $b_1(X)$.