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Show that for the de Rham cohomology, $H^{1}_{dR}({T^{2}})$ is isomorphic to $\mathbb{R}^{2}$ by showing that the following map: $[\alpha] \to (\int_{S^{1}}f^{*}_{1}\alpha,\int_{S^{1}}f^{*}_{1}\alpha)$, where $f_{1}=(\theta,c_{1}), f_{2}=(c_{2},\theta)$ for $c_{1},c_{2}$ constants (seen as maps $S^{1} \to S^{1} \times S^{1}$) is an isomorphism.

I have shown that the map is well-defined, linear and independent of the constants chosen. However, I'm having trouble with showing that it is surjective and injective. Surjectivity seems to be easy to show, but I'm not sure on how to go about it. For injectivity, it suffices to show that if the integrals are $0$, then $\alpha$ is an exact form, yet I'm having troubles with it as well. What would be a good way to tackle this problem?

betelgeuse
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    Do you already know the map $H^1_{dR}(S^1)\rightarrow \mathbb{R}$ given by $[\alpha]\mapsto \int_{S^1} \alpha$ is an isomorphism? If not, I would start there. – Jason DeVito - on hiatus Nov 18 '18 at 16:56
  • Yes, I do know that. I'm not sure on how to use this fact to show that the map $H^{1}{dR}(T^{2}) \to \mathbb{R}^{2}$ is an isomorphism though. Would perhaps composing the induced cohomology map $F{1}^{}$ by $f_{1}^{}$ with the isomorphism $H^{1}_{dR}(S^{1}) \to \mathbb{R}$ be a good way to proceed? – betelgeuse Nov 18 '18 at 21:28
  • Sorry for the delay, been doing holiday traveling. (Also, if you want to message me, you should precede your comment with "@Jason" or something like that, so that I get pinged). This gives you a way of showing surjectivity more or less immediately. Given $(r,0)\in \mathbb{R}^2$, by surjectivity of $H^1_{dR}(S^1)\rightarrow \mathbb{R}$, there is a $1$-form $\beta$ on $S^1$ with $\int_{S^1} \beta = r$. Now, let $\alpha(\theta_1, \theta_2) = \beta(\theta_1)$ . Then $[\alpha]\mapsto (r,0)$. To get $(0,r)$, use the same trick. To get $(r_1,r_2)$.... – Jason DeVito - on hiatus Nov 21 '18 at 16:32
  • use the fact that $H^1_{dR})(T^2)\subseteq \mathbb{R}^2$ is a subspace. It does seem like injectivity is trickier. You may need to do something like what Pedro did below. – Jason DeVito - on hiatus Nov 21 '18 at 16:33

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Recall that $H_1(T^2)$ is $\mathbb Z^2$, and is spanned by the class of the simplices $f_1$ and $f_2$, and in fact your map is coming from a pairing $\Omega^*(T^2) \otimes C_*(T^2,\mathbb R)\to\mathbb R$ such that $\langle \omega,\sigma\rangle = \int_\sigma \omega$. This is the same as the corresponding map $\Omega^*(T^2)\to C^*(T^2,\mathbb R)$, and what your question is claiming is that this is an isomorphism on cohomology, that is, a quasi-isomorphism. One can actually consider this map for any manifold $X$, let us call it $r_X$, the de Rham pairing.

Now something remarkable about $r_X$ this is that it is compatible with, for example, the Mayer-Vietoris sequence, in the sense that if $U,V$ are open and cover $X$, then the following diagram commutes

$$\require{AMScd} \begin{CD} 0@>>>\Omega^*(X) @>>> \Omega^*(U)\oplus \Omega^*(V) @>>> \Omega(U\cap V) @>>> 0\\ {} @VVr_XV @VV{r_U\oplus r_V} V @VV{r_{U\cap V}}V\\ 0 @>>> C^*(X) @>>> C^*(U)\oplus C^*(V) @>>> C^*(U\cap V) @>>> 0 \end{CD} $$

It follows that if $r_X$ is a quasi-isomorphism for $U,V$ and $U\cap V$, it is an isomorphism. Indeed, this is a statement of homological algebra and follows from the so-called 5 lemma applied to the "infinite staircase" diagram obtained by lookin at the long exact sequences of the two Mayer-Vietoris sequences above.

Now you can cover the torus $T^2$ by two open sets $U$ and $V$ which are homotopic to the $1$-sphere, and whose intersection is a disjoint union of two $1$-spheres. Hence, proving it for the $1$-sphere suffices.

In this case you know that $H^*_{dR}(S^1)$ is spanned by $1$ and an angle form $d\theta$, and that $H_*(S^1)$ is spanned by, more or less, the identity $S^1\to S^1$. Then the induced by on homology is simply the isomorphism $\mathbb R\otimes\mathbb R\to\mathbb R$, so you get what you wanted.

Pedro
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