A local trivial fibration If M is compact and connected manifold. Let $\pi:M\longrightarrow T^{2} $ be a local trivial fibration. 
Why $\pi^{*}:H^{1}(T^{2})\longrightarrow H^{1}(M)$ is injective?
 A: (I will use "connected" to mean "path-connected" throughout)
If it is a fibration, then you have a long exact sequence of homotopy groups 
$\pi_2(T^2)\to \pi_1(F)\to \pi_1(M)\to \pi_1(T^2)\to \pi_0(F)\to \pi_0(M)$, where $F$ is the fiber. 
As $M$ is connected and $\pi_2(T^2) = 0$ you get $0\to \pi_1(F)\to \pi_1(M)\to \pi_1(T^2)\to \pi_0(F)\to \{*\}$
Now if $F$ is connected, $\pi_1(M)\to \pi_1(T^2)$ is surjective so by the Hurewicz theorem, $H_1(M)\to H_1(T^2)$ is as well, so by the universal coefficient theorem $H^1(T^2)\to H^1(M)$ is injective. 
If $F$ isn't connected, you have to see that $\pi_0(F)$ is finite, so that in fact the image of $\pi_1(M)$ in $\pi_1(T^2)$ has finite index, therefore it is the same with $H_1(M)$ inside $H_1(T^2)$. Passing to the universal coefficient theorem you have a map $\hom(H_1(T^2),\mathbb Z)\to \hom(H_1(M),\mathbb Z)$; let $f:H_1(T^2)\to \mathbb Z$ and assume it becomes zero when you go to $H_1(M)\to \mathbb Z$. 
Then $f$ factors through $H_1(T^2)/\mathrm{im}(H_1(M))$ which is finite by what we said earlier, hence torsion : it must be the zero map. It follows that $\hom(H_1(T^2),\mathbb Z)\to \hom(H_1(M),\mathbb Z)$ is injective, so by the UCT, $H^1(T^2)\to H^1(M)$ is injective. 
Now we have to see why $\pi_0(F)$ is always finite. This follows from the following consideration : cover $T^2$ by (finitely many, by compactness) connected opens $U_i$ such that the fibration is trivial over them, so it looks like $U_i\times F\to U_i$.
Note that $\pi^{-1}(U_i)$ is open ($U_i$ is), therefore it is locally connected : so is $U_i\times F$. It follows that if $C$ is a connected component of $F$, it is open in $F$ (let $x\in F$ and fix some $t\in U_i$. Then there is some neighbourhood $W$ of $(t,x)$ in $U_i\times F$ that is path-connected, by local path-connectedness of $U_i\times F$. Now $pr_F(W)$ is open, is connected, and contains $x$; so $x$ is in the interior of its connected component - we can actually prove that way that $F$ is locally connected).
Since $F$ is closed (as $\pi^{-1}(*)$) in a compact manifold $M$, it is also compact. Therefore as it is covered by its connected components, which are open, a finite number of them suffices, and so there's only finitely many of them : $\pi_0(F)$ is finite.
