# 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?

• I would guess on using Hurewicz' theorem, in particular since $\mathrm{Hom}(\pi_1(S,s_0), (\mathbb{R},+))$ is essentially the abelization of the fundamental group. – Peter Wildemann May 12 '18 at 8:25

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.

• Your fact is wrong. You can produce easy counterexamples. It is only true that primitive curves can be represented by embeddings, but this does not really mess with your proof. – ThorbenK May 13 '18 at 13:33
• @ThorbenK, thank you for the correction! I'm not really sure what a primitive curve is (I'm an algebraic topologist, and my knowledge of differential topology is pretty primitive). How should I correct my statement? – Tyrone May 13 '18 at 13:46
• It has to be a primitive element in $\pi_1(X)$ i.e. not a non-trivial multiple of some curve. – ThorbenK May 13 '18 at 14:01