Pull-back of 1-form

Let $f : \mathbb{R} \rightarrow \mathbb{S}^1$ be the parameterization $f(\theta) = (\cos\theta, \sin\theta)$. If $\alpha$ is a 1-form on $\mathbb{S}^1$, prove that $$\int_{\mathbb{S}^1} \alpha = \int_0^{2\pi} f^*\alpha.$$ I know $$f^*\alpha(\theta) = (df_\theta)^*\alpha(f(\theta)),$$ but I'm not sure how to proceed.

• Proving the integral equation – user453280 Jun 7 '17 at 22:51
• How did you define the integral of a 1-form on $\mathbb{S}^1$? – Willie Wong Jun 8 '17 at 4:54
• I believe $\int_{\mathbb{S}^1} \alpha = \int_0^{2\pi} \alpha(f'(\theta))\,d\theta$ – user453280 Jun 8 '17 at 5:04
• So you are identifying $\mathbb{S}^1$ with its canonical embedding in $\mathbb{R}^2$, and implicitly lifting $\alpha$ to a one-form on $\mathbb{R}^2$? Then does your identity not follow from just pushing symbols around and rearranging stuff? – Willie Wong Jun 8 '17 at 13:38