$S^1$-valued function on $T^n$

$\mathbb{R}$ and $S^1$ have isomorphic lie algebras thus we can identify forms with values in $TS^1$ with forms with values in $\mathbb{R}$ as follows. We take angle functions $$\theta_{\alpha}: U_{\alpha}\subsetneq S^1 \to \mathbb{R}$$ given by $e^{i2\pi \theta}\mapsto \theta$. Those maps gives $S^1$ a smooth chart. We note that $\theta_\alpha \circ f$ is in general not globally well defined on $T^n$, we can patch together a global 1-form defined locally on $f^{-1}(U_\beta)$ by $d(\theta_\beta \circ f)$ by noticing that $d\theta_{\alpha}$ and $d\theta_\beta$ agree on $U_\beta \cap U_\alpha$ and do not depend on the branch of the logarithm we choose. This is definitely a closed 1 from as it is defined locally as the exterior derivative of a smooth function.

Let $[df] \in H_{dr}^1(T^n,\mathbb{R})$ be the class represented by the 1-form above. We have a natural inclusion of the chain complex that defines $H^1(T^n,\mathbb{Z})$ in the chain complex that defines $H^1(T^n,\mathbb{R})$ using the inclusion of $\mathbb{Z}\hookrightarrow \mathbb{R}$. We will use de-rham isomorphism to identify $[df] $ with an element of $H^1(T^n,\mathbb{R})$. Recall that the isomorphism is given by $[df]\mapsto (\sigma \mapsto \int_{\sigma}df)$. We can further identify $[df]$ with an element of $H^1(T^n,\mathbb{Z})$. In order to see this, we take a smooth chain $\sigma:[0,1] \to T^n$ representing one of the generators of $H_1(T^n,\mathbb{Z})$. We will now compute $$\int_{\sigma}df$$ and show that its an integer. We note that $f \circ \sigma:[0,1] \to S^1$ is a map from a simply connected domain. Hence we can lift it $\widetilde{f \circ \sigma}:[0,1]\to \mathbb{R}$ to a map that satisfy $$e^{i2\pi \widetilde{f \circ \sigma}}= f\circ \sigma.$$ We deduce that
$$\int_{\sigma}df=\int_0^1 d\widetilde{f \circ \sigma}=\widetilde{f \circ \sigma}(1)-\widetilde{f \circ \sigma}(0).$$ As $\sigma$ is closed we have $\sigma(0)=\sigma(1)$. Thus we have $$e^{i2\pi \widetilde{f \circ \sigma(1)}}=e^{i2\pi \widetilde{f \circ \sigma(0)}},$$ and this implies that $\widetilde{f \circ \sigma}(1)-\widetilde{f \circ \sigma}(0)\in \mathbb{Z}$. If you have some experience with complex analysis you have probably noticed that $$\int_{\sigma}df=\frac{1}{i2\pi}\int_{0}^1d \ln(f \circ \sigma)$$
which is the winding number of $f \circ \sigma$ and this is the missing connection with degree theory.