[proofplan] We replace a neighbourhood of $N$ in $M$ by a fibrewise star-shaped tubular neighbourhood in the normal bundle of $N$. On this tubular neighbourhood, the radial contraction onto the zero section gives a homotopy operator $K$ for differential forms. The homotopy formula expresses a form as the sum of its pullback from the zero section and an exact term, up to the [exterior derivative](/theorems/1525) of the form. Since $\alpha$ is closed and restricts to zero on $N$, this formula produces the desired primitive, and the radial vector field vanishes on the zero section, giving the relative condition on $\beta$. [/proofplan]

[step:Choose a fibrewise star-shaped tubular neighbourhood inside $V$] Choose a [smooth vector bundle](/page/Smooth%20Vector%20Bundle) $\nu\to N$ complementary to $TN$ in $TM|_N$, and let $\pi_\nu:\nu\to N$ denote its bundle projection. By the [Tubular Neighbourhood Theorem](/theorems/2276) for closed embedded submanifolds, there are an open neighbourhood $E_0\subset\nu$ of the zero section and a smooth embedding \begin{align*} \Phi:E_0\to M \end{align*} whose image is an open neighbourhood of $N$ in $M$ and whose restriction to the zero section is the inclusion $i:N\to M$. Since $V\subset M$ is open and contains $N$, we shrink $E_0$ to a fibrewise star-shaped open neighbourhood $E\subset E_0$ of the zero section such that $\Phi(E)\subset V$. Concretely, after choosing a smooth bundle metric on $\nu$, one may take \begin{align*} E=\{v\in\nu: |v|_{\nu}<\rho(\pi_\nu(v))\} \end{align*} for a positive smooth function $\rho:N\to(0,\infty)$ chosen small enough that this set lies in $\Phi^{-1}(V)$. This choice is possible because $N$ is paracompact as a closed embedded submanifold of the second-countable Hausdorff manifold $M$. Define \begin{align*} U:=\Phi(E)\subset V. \end{align*} Then $U$ is an open neighbourhood of $N$ in $V$, and $\Phi:E\to U$ is a diffeomorphism. Let \begin{align*} z:N\to E \end{align*} be the zero-section inclusion. Then $\Phi\circ z=i_U$ and $\pi_\nu\circ z=\operatorname{id}_N$. [/step]

[step:Construct the radial homotopy operator on the tubular neighbourhood]Define the radial contraction \begin{align*} H:[0,1]\times E\to E,\qquad H(t,v)=t v. \end{align*} The fibrewise star-shaped choice of $E$ ensures that $t v\in E$ for every $v\in E$ and every $t\in[0,1]$. Write $H_t:E\to E$ for the smooth map $H_t(v)=t v$. Thus $H_1=\operatorname{id}_E$ and $H_0=z\circ\pi_\nu$. For $\gamma\in\Omega^r(E)$ with $r\ge 1$, define the homotopy operator \begin{align*} K\gamma\in\Omega^{r-1}(E) \end{align*} by \begin{align*} (K\gamma)_v(w_1,\dots,w_{r-1})=\int_0^1 \gamma_{H_t(v)}\bigl(\partial_tH(t,v),d(H_t)_v(w_1),\dots,d(H_t)_v(w_{r-1})\bigr)\,d\mathcal L^1(t). \end{align*} Here $v\in E$, the vectors $w_1,\dots,w_{r-1}$ lie in $T_vE$, and $\partial_tH(t,v)\in T_{H_t(v)}E$ denotes the velocity of the curve $t\mapsto H_t(v)$. For $r=0$, set $K\gamma=0$. The homotopy formula for differential forms gives, for every $\gamma\in\Omega^r(E)$, \begin{align*} H_1^*\gamma-H_0^*\gamma=dK\gamma+K(d\gamma). \end{align*} Since $H_1=\operatorname{id}_E$ and $H_0=z\circ\pi_\nu$, this becomes \begin{align*} \gamma-\pi_\nu^*z^*\gamma=dK\gamma+K(d\gamma). \end{align*}[/step]

[guided]The point of passing to the tubular neighbourhood is that there is now a canonical deformation back to $N$: each normal vector $v\in\nu$ can be scaled to $t v$. The fibrewise star-shaped condition on $E$ is exactly what makes the map \begin{align*} H:[0,1]\times E\to E,\qquad H(t,v)=t v \end{align*} well-defined on the whole interval $[0,1]$. For each $t\in[0,1]$, define \begin{align*} H_t:E\to E,\qquad H_t(v)=t v. \end{align*} Then $H_1=\operatorname{id}_E$, while $H_0$ sends every vector $v\in E$ to the zero vector in the fibre over $\pi_\nu(v)$. Therefore \begin{align*} H_0=z\circ\pi_\nu. \end{align*} The homotopy operator lowers degree by one. If $\gamma\in\Omega^r(E)$ with $r\ge 1$, define $K\gamma\in\Omega^{r-1}(E)$ by \begin{align*} (K\gamma)_v(w_1,\dots,w_{r-1})=\int_0^1 \gamma_{H_t(v)}\bigl(\partial_tH(t,v),d(H_t)_v(w_1),\dots,d(H_t)_v(w_{r-1})\bigr)\,d\mathcal L^1(t). \end{align*} The vector $\partial_tH(t,v)$ is inserted into the first slot of $\gamma$, and the remaining vectors are pushed forward by $d(H_t)_v$. The integral is taken with respect to one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) $\mathcal L^1$ on $[0,1]$. For $r=0$, we set $K\gamma=0$ because there is no form of degree $-1$. The homotopy formula for differential forms states that this operator satisfies \begin{align*} H_1^*\gamma-H_0^*\gamma=dK\gamma+K(d\gamma). \end{align*} Substituting $H_1=\operatorname{id}_E$ and $H_0=z\circ\pi_\nu$ gives \begin{align*} \gamma-\pi_\nu^*z^*\gamma=dK\gamma+K(d\gamma). \end{align*} This identity is the algebraic heart of the relative Poincare lemma: it says that a form differs from its restriction to the zero section by an exact term, with an additional error term involving its exterior derivative.[/guided]

[step:Apply the homotopy formula to the pulled back closed form] Define \begin{align*} \widetilde{\alpha}:=\Phi^*(\alpha|_U)\in\Omega^k(E). \end{align*} Since exterior differentiation commutes with pullback and $d\alpha=0$ on $V$, \begin{align*} d\widetilde{\alpha}=d\Phi^*(\alpha|_U)=\Phi^*(d(\alpha|_U))=0. \end{align*} Also, because $\Phi\circ z=i_U$ and $i_U^*(\alpha|_U)=i_V^*\alpha=0$, \begin{align*} z^*\widetilde{\alpha}=z^*\Phi^*(\alpha|_U)=(\Phi\circ z)^*(\alpha|_U)=i_U^*(\alpha|_U)=0. \end{align*} Applying the homotopy formula to $\gamma=\widetilde{\alpha}$ gives \begin{align*} \widetilde{\alpha}-\pi_\nu^*z^*\widetilde{\alpha}=dK\widetilde{\alpha}+K(d\widetilde{\alpha}). \end{align*} The two terms involving $z^*\widetilde{\alpha}$ and $d\widetilde{\alpha}$ vanish, so \begin{align*} \widetilde{\alpha}=dK\widetilde{\alpha}. \end{align*} [/step]

[step:Push the primitive back to $U$ and verify it differentiates to $\alpha$] Define \begin{align*} \beta:=(\Phi^{-1})^*(K\widetilde{\alpha})\in\Omega^{k-1}(U). \end{align*} Since exterior differentiation commutes with pullback by the diffeomorphism $\Phi^{-1}:U\to E$, \begin{align*} d\beta=d(\Phi^{-1})^*(K\widetilde{\alpha})=(\Phi^{-1})^*(dK\widetilde{\alpha})=(\Phi^{-1})^*\widetilde{\alpha}. \end{align*} By the definition of $\widetilde{\alpha}$, \begin{align*} (\Phi^{-1})^*\widetilde{\alpha}=(\Phi^{-1})^*\Phi^*(\alpha|_U)=\alpha|_U. \end{align*} Hence \begin{align*} d\beta=\alpha|_U. \end{align*} [/step]

[step:Check that the primitive vanishes along the submanifold] It remains to verify the relative condition. Since $i_U=\Phi\circ z$, \begin{align*} i_U^*\beta=z^*\Phi^*(\Phi^{-1})^*(K\widetilde{\alpha})=z^*(K\widetilde{\alpha}). \end{align*} For every $x\in N$, the path $t\mapsto H_t(z(x))$ is constant equal to $z(x)$, so \begin{align*} \partial_tH(t,z(x))=0\in T_{z(x)}E. \end{align*} The defining formula for $K$ therefore gives \begin{align*} (z^*K\widetilde{\alpha})_x=0 \end{align*} as a $(k-1)$-form on $T_xN$. When $k=1$, this says the resulting smooth function has value $0$ at every point of $N$. Thus \begin{align*} i_U^*\beta=0. \end{align*} The form $\beta\in\Omega^{k-1}(U)$ therefore satisfies both required conclusions. [/step]