Im confused about the proof of Poincaré's Lemma given in the book Geometry, Topology and Physics by M. Nakahara. He states that for a closed $r$-form $\omega $ on a contractible chart U the composite map $f_1^*\circ F^*\omega$ is zero, where $F$ is a smooth map $F:U\times I \rightarrow U$ such that
$$
F(x,0)=x,\;\;\;\;\; F(x,1)=p_0 \;\;\;\;\;\text{for}\; x\in U
$$
and the map $f_t:U\rightarrow U\times I$ is defined as $f_t(x)=(x,t)$. He argues that this is true due to the fact that $F\circ f_1:U\rightarrow U$ is a constant map $x\mapsto p_0$, hence $(F\circ f_1)^*=0$. I can't see why the pullback of a constant map should be zero. I haven't worked with pullbacks for very long so maybe im getting confused as to how they work, but as far as I understand; the map $f_1^* \circ F^*$ you'd start with the points $x\in U$ and map them to points $(x,t)\in U\times I$. This was the $F^*$ part, which seems clear to me, however using $f_1^*$ on these points gets me confused. Using the pullback $f_1^*$ on these points would only map the points for which $t=1$ i.e $(x,1)\in U\times I $ to $x\in U$ and even then the map becomes the idenitity $x\mapsto x$. What am I misunderstanding?
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If this is confusing I'll provide some context:
In the proof he starts out stating the lemma:
If a coordinate Neightbourhood U of a manifold M is contractible to a point $p_0\in M$, any closed $r$-form on U is also exact.
In the proof he assumes U is smoothly contractible to $p_0$ in the sense that there exists a smooth map $F$ as defined above
He then consideres an $r$-form $\eta\in \Omega^r(U\times I)$
$$
\eta = a_{{i_1}…i_r}(x,t) dx^{i_1} \wedge … \wedge dx^{i_r}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;+b_{{j_1}…j_{r-1}}(x,t) dt\wedge dx^{j_1} \wedge … \wedge dx^{j_{r-1}}
$$
where $x$ is the coordinate of U and t of I.
Next he defines the map $P:\Omega ^r(U\times I ) \rightarrow \Omega^{r-1}(U)$ by
$$
P\eta \equiv\left(\int_0^1 b_{{j_1}…j_{r-1}}(x,s) \right)dx^{j_1} \wedge … \wedge dx^{j_{r-1}}
$$
The last map he defines is $f_t$ as defined above and explicitly gives
$$
f^*_t\eta=a_{{i_1}…i_r(x,t) dx^{i_1}} \wedge … \wedge dx^{i_r}\in\Omega ^r(U)
$$
To prove Poincaré's lemma he says that if he replaces $\eta$ by $F^*\omega\in \Omega ^r(U\times I)$, where $\omega $ is a closed $r$-form on $U$, in
$$
d(P\eta)+P(d\eta)=f_1^*\eta-f_0^*\eta\\
\downarrow\\
d(PF^*\omega)+P(dF^*\omega)=f_1^*\circ F^*\omega-f_0^*\circ F^*\omega\\
$$
he can show that $\omega = d(-PF^*\omega)$, which proves poincaré's lemma.
He argues that the first term vanishes since $F\circ f_1:U\rightarrow U$ is a constant map $x\mapsto p_0$, hence $(F \circ f_1 )^* = 0$
Best Answer
You have the smooth map $F:U\times I\to U$, with $U\subset M$ local chart and $I\subset \mathbb R$ s.t.
$F(x,0)=x$, $F(x,1)=p_0$ for $x\in U$. Consider then $\{f_t:U\to U\times I|f_t(x)=(x,t)\}_{t\in I}$.
For $\omega\in\Omega^r(U)$ (I'll write the form as $\underset{I=(i_1,\dots, i_r) \text { s.t. }1\le i_1<\dots<i_r\le \dim(M)}{\sum}\omega_{i_1\dots i_r} dx^{i_1}\wedge\dots \wedge dx^{i_r}$, where $\omega_{i_1\dots i_r}$ are smooth functions and $x^{i_1},\dots,x^{i_r}$ are the local coordinates of the open set $U$), we can now compue the pullback $f_1^*\circ F^* \omega$: $$f_1^*\circ F^*(\omega)=f_1^*(F^*(\omega))=(F\circ f_1)^*(\omega)=(F\circ f_1)^*\bigg(\sum_I\omega_{i_1\dots i_r}dx^{i_1}\wedge\dots\wedge dx^{i_r}\bigg)=\\=\sum_I(\omega_{i_1\dots i_r}\circ F\circ f_1) d(x^{i_1}\circ F\circ f_1)\wedge\dots\wedge d(x^{i_r}\circ F\circ f_1)=\\=\sum_I \omega_{i_1\dots i_r}(F(f_1(x)))d(x^{i_1}\circ F\circ f_1)\wedge\dots\wedge d(x^{i_r}\circ F\circ f_1)=\\ =\sum_I \omega_{i_1\dots i_r}(F(x,1))d(x^{i_1}(F(x,1)))\wedge\dots\wedge d(x^{i_r}(F(x,1)))=\\=\sum_I\omega_{i_1\dots i_r}(p_0)d(x^{i_1}(p_0))\wedge\dots\wedge d(x^{i_r}(p_0))$$ now observe that $F(x,1)=p_0\in U$,so $x^{i_k}(\underbrace{p_0}_{\in U})=p_0\in\mathbb R$ and the exterior derivative of $x^{i_k}(p_0)$ is zero for each $k.$