Pullback of a function

Question

I take the following example from the book of Introduction to Manifolds by Loring Tu page 198 example 17.16

Let $h : \mathbb{R} \rightarrow S^1 \subset \mathbb{R}^2$ be given by $h(t) = (x,y)=(\cos t,\sin t)$.

Why this is true: $h^*x = \cos t $ and $h^*y = \sin t $ ?

I'm a little confused, I tried to apply the definition of the pullback of a function but this doesn't give me the answer in the book!

Answer 1

The following definition is from the book you mention but I couldn't find your example there.

If $\omega$ is a 1 -form on $M,$ we define its pullback $F^{*} \omega$ to be the 1 -form on $N$ given by $$ \left(F^{*} \omega\right)_{p}\left(X_{p}\right)=\omega_{F(p)}\left(F_{*} X_{p}\right) $$ for any $p \in N$ and $X_{p} \in T_{p} N$

You need to be certain of your definitions and objects you are working with (in this case your $x$ and $y$, pullback). An example calculation:

Consider the one-forms $\alpha_{(x_1,x_2)}=x_1dx_1+0dx_2$ and $\beta_{(x_1,x_2)}=0dx_1+x_2dx_2$. Let $h: \mathbb{R} \rightarrow \mathbb{R}^2$ be given by $h(t)=(\cos(t),\sin(t))$. Consider a general vector field $X$ on $\mathbb{R}$, $X=X^1 \frac{\partial}{\partial t}$. Jacobian matrix of $h_*$ is $(h_*)_{(t)}= \big(\begin{smallmatrix} -\sin(t) \\ \cos(t) \end{smallmatrix}\big).$ Hence $(h_*)_{(t)}X = -\sin(t)X^1\frac{\partial}{\partial x_1}+\cos(t)X^1\frac{\partial}{\partial x_2}.$

So $$\alpha_{h(t)}(h_*X)=\alpha_{(\cos(t),\sin(t))}(h_*X)=-\cos(t)\sin(t)X^1$$ $$\beta_{h(t)}(h_*X)=\beta_{(\cos(t),\sin(t))}(h_*X)=\sin(t)\cos(t)X^1$$ by comparing coefficients $$h^*\alpha = -\cos(t)\sin(t)dt \text{ and } h^*\beta = \sin(t)\cos(t)dt.$$