de-Rham cohomology $H^1(S)$ and $ \operatorname{Hom}(\pi_1(S, s_0), (\mathbb{R},+))$ are isomorphic? Let $S$ be a connected smooth 2-dimensional manifold.
Let $H^1(S)$ be its first De-rham cohomology group.
\begin{align*}\operatorname{Hom}(\pi_1(S, s_0), (\mathbb{R},+))=\{f \vert f :\pi_1(S, s_0)\to (\mathbb{R},+)\text{ is a group homomorphism} \}\end{align*}
Define\begin{align*} \Phi:H^1(S)\to \operatorname{Hom}(\pi_1(S, s_0), (\mathbb{R},+))  \end{align*}
as $ \Phi( [\alpha])([c])=\int_c \alpha $
I know that $\Phi$ is an injective linear map.
How can I show that it is surjective?
 A: It is a fact from differential topology that any primitive curve $c:S^1\rightarrow S$ is homotopic to an embedding. It follows that we may represent each generating class $[c_i]\in\pi_1S$ by an embedded circle $C_i\subset S$, and moreover we may assume that any two such $C_i$, $C_j$ have transverse intersection.
Now let $\theta:\pi_1S\rightarrow\mathbb{R}$ be a homomorphism. Then this will be completely determined by its action on chosen generators, so it will suffice in the following to work only with these classes $[c_i]$.
Now for each generator $c_i:S^1\rightarrow S$ in $\pi_1S$ we can quite easily construct a 1-form $\alpha'_i\in\Lambda^1S^1$ such that
$$\int_{S^1}\alpha'_i=\theta(c_i).$$
If we choose suitable tubular neighbourhoods $T_i$ for each embedded circle $C_i$ then we can extened each $\alpha_i'$ first over $T_i$, and then by use of bump functions to all of $S$. Everything works out, since any curves that intersect non-trivially do so transversely, $C_i\pitchfork C_j$, so in such a case we can chose $T_i$, $T_j$ suitably and define $\alpha'_i$, $\alpha'_j$ and their extensions in such a way that everything agrees over intersections.
The end result is that we can glue the family $\alpha_i'$ together to give a globally defined 1-form $\alpha\in \Lambda^1S$ which satisfies $c_i^*\alpha=\alpha|_{C_i}=\alpha'_i$.
Now we have for each generating class $[c_i]\in\pi_1S$ that
$\Phi(\alpha,[c_i])=\int_{C_i}\alpha=\int_{S^1}c^*\alpha=\int_{S^1}\alpha'_i=\theta(c_i)$.
Hence the homomorphism $\theta$ lies in the image of $\Phi$, which is therefore surjective.
Edit: A small error has been corrected after being observed in the comments.
