Smooth functions and pullback map This question was asked in my assignment on manifolds and I am struck on it.

Question: For a 1-form  w= f(x) dx on $\mathbb{R}$  , define  $\int_{a}^{b} w  = \int_{a}^{b} f(x) dx $ for $a, b \in \mathbb{R}$.

(a) Let h be a smooth function on $\mathbb{R}$, show that $\int_{a}^{b} dh =h(b) -h(a)$.
(b) Let $U\subset \mathbb{R}^2 -${0} be an open set and let $\alpha : \mathbb{R} \to U $  be a smooth map. Show that $\alpha^{*}(dg) = d( g\circ \alpha)$ for a smooth function g on U.
(c) Consider the map $\alpha : \mathbb{R}  \to \mathbb{R}^2$ given by $\alpha (t) = (cos(2 \pi t) , sin (2\pi t))$. Consider the 1-form $n= \frac{x} { (x^2 +y^2)^{1/2}} dy - \frac{y} { (x^2 +y^2)^{1/2}}$ on $\mathbb{R}^2 -${0}. Compute $\int_{0}^{1} \alpha^{*} (n)$.
(d) Show that $dn=0$ but there exists no smooth  f on $\mathbb{R}^2$ such that $n=df$.
Attempt: I have proved (a).
In (b) and (c) the problem I am facing is that I am not very confident in solving the problems of Pullback maps. Pullbacks are defined as follows: Let $f : M \to N$ be a smooth map. If $ w\in \Omega^k(N)$, we get $f^{*}(w) \in \Omega^{k} (M)$. I have studied some relevant properties of this map. But still I am not confident in using the properties of this map in questions.
(d) I have proved that $dn$ will be equal to 0 and let on the contrary there exists a smooth function such that $n=df$ but I am not able to proceed towards a contradiction.
Can you please give a few hints for this problem?
Thanks!
 A: $(b).$ Here it depends by what kind of definition you are about $d$ and pullback operation. For example for me $dg$ is point wise defined like:
$$dg_q:=\frac{\partial g}{\partial x}(q)dx_q+ \frac{\partial g}{\partial y}(q)dy_q$$
Regarding the pullback operation, given a smooth function $\alpha\colon M\to N$, and a $1$-form (for example, but you can generalise for other spaces) $\omega\in \Omega^1(N)$, then
$\alpha^*(\omega)\in \Omega^1(M)$ and
$$\alpha^*(\omega)_p(v_p):=\omega_{\alpha(p)}(\alpha^*v_p)$$
where $v_p\in T_pM$ and
$$\alpha^*v_p(f):=v_p(f\circ \alpha)$$
Thus you get a smooth map $\alpha^*\colon \Omega^1(N)\to \Omega^1(M)$ that is point wise $\alpha^*_p\colon \Omega^1_{\alpha(p)}(N)\to \Omega^1_p(M)$ a linear map of vector spaces. Thus the map $\alpha^*$ results to be a morphism of vectors bundles (this is just an information, we don’t need to know for our purpose).
Now we can try to solve your problem. For any point $p\in M=\mathbb{R}$ we have
$$\alpha^*(dg)_p(\frac{\partial}{\partial t}_p)=(dg)_{\alpha(p)=(\alpha_1(p),\alpha_2(p))}(\alpha^* \frac{\partial}{\partial t}_p )= (dg)_{\alpha(p)}(\frac{\partial \alpha_1}{\partial t}_p\frac{\partial}{\partial x}_p+\frac{\partial \alpha_2}{\partial dt}\frac{\partial}{\partial y}_p)= \frac{\partial g}{\partial x}(\alpha(p)) \frac{\partial \alpha_1}{\partial t}_p + \frac{\partial g}{\partial y}(\alpha(p)) \frac{\partial \alpha_2}{\partial t}_p =d(g\circ \alpha)_p(\frac{\partial}{\partial t}_p )$$
This holds for any $p\in M$, so $\alpha^*(dg)=d(g\circ \alpha)$.
$(c).$ Let’s compute $\alpha^*(n)$:
$$\alpha^*(n)= \frac{cos(2\pi t)} { (1)^{1/2}} d(sin(2\pi t)- \frac{sin(2\pi t)} { (1)^{1/2}}d(cos(2\pi t)=2\pi\left( cos^2(2\pi t)+sin^2(2\pi t)\right)dt=2\pi dt$$
This means that
$$\int_0^1 \alpha^*(n)=2\pi \int_0^1 dt=2\pi$$
$(d).$ This is the more interesting one, I think. You can proceed in this way. Suppose by contradiction that there exists a smooth function $f$ of $\mathbb{R}^2$ such that $n=df$. Then this means
$$\int_0^1 \alpha^*(n)=\int_0^1 d(f\circ \alpha)=(f\circ \alpha)(1)-(f\circ \alpha)(0)=f(1,0)-f(1,0)=0$$
because of points $(b).$ and $(a).$
Of course this contradicts the fact that $\int_0^1 \alpha^*(n)=2\pi$, by point $(c).$
Remark: You’ve just proved that $H^1_{dR}(\mathbb{R}^2\setminus \{0\})$ is non-trivial.
