This is a problem from Jänich's Vector Analysis:
Prove directly from the definition that $[\omega] \rightarrow \int_{S^1}\omega$ defines an isomorphism $H^1(S^1)\cong \mathbb{R}$, and go on to show that $\text{dim}H^1(S^1 \times S^1) \geq 2$.
It's easy to see that the map is linear. I am not sure what definition he means, but I'm guessing it's the definiton of $$H^1(S^1):=\frac{\text{ker}(\text{d}:\Omega^1S^1\rightarrow \Omega^2S^1)}{\text{im}(\text{d}:\Omega^0S^1\rightarrow \Omega^1S^1)}$$
I know that $\text{ker}(\text{d}:\Omega^1S^1\rightarrow \Omega^2S^1)=\Omega^1S^1$, but I don't know what $\text{im}(\text{d}:\Omega^0S^1\rightarrow \Omega^1S^1)$ is. So I try to show that the map is injective and surjective. To see that it is injective I need to show that $[\omega]=[\alpha]$ if $\int_{S^1}\omega=\int_{S^1}\alpha$, but I can't see how. And surjectivity seems harder (as always), how can I contruct a form such that $\int_{S^1}\omega = c$ for $c \in \mathbb{R}$?
For the second part the book suggests to use the functoriality of $H^1$, to get:
$$H^1(S^1)\rightarrow H^1(S^1 \times S^1) \rightarrow H^1(S^1)$$ with projection and inclusion induced maps and $\text{dim}H^1(S^1)=1$.
how can I conclude from there that $\text{dim}H^1(S^1 \times S^1) \geq 2$?