[proofplan] We prove that the construction produces a closed global form, then check the two possible choices in the definition. First, changing the lift of a fixed closed representative changes the resulting global $(k+1)$-form by an exact form on $M$. Second, replacing the representative by a cohomologous one again changes the resulting global form by an exact form. Finally, linearity follows by choosing compatible lifts and using linearity of the maps $s_k$, $r_{k+1}$, and $d$. [/proofplan]

[step:Verify that the construction produces a closed global form]Let $\gamma \in \Omega^k(U \cap V)$ be closed, so $d\gamma=0$, and let $(\alpha,\beta) \in \Omega^k(U) \oplus \Omega^k(V)$ satisfy $s_k(\alpha,\beta)=\gamma$. Define the pair \begin{align*} (a,b) := (d\alpha,d\beta) \in \Omega^{k+1}(U) \oplus \Omega^{k+1}(V). \end{align*} Because exterior differentiation commutes with restriction, \begin{align*} s_{k+1}(a,b) &= d\beta|_{U \cap V} - d\alpha|_{U \cap V} \\ &= d(\beta|_{U \cap V} - \alpha|_{U \cap V}) \\ &= d(s_k(\alpha,\beta)) \\ &= d\gamma \\ &= 0. \end{align*} Thus $(a,b) \in \ker s_{k+1}$. By exactness at $\Omega^{k+1}(U)\oplus \Omega^{k+1}(V)$, there exists $\eta \in \Omega^{k+1}(M)$ such that \begin{align*} r_{k+1}(\eta)=(d\alpha,d\beta). \end{align*} We now show that $\eta$ is closed. Since $r_{k+2}$ is injective by exactness at $\Omega^{k+2}(M)$, it suffices to prove $r_{k+2}(d\eta)=0$. Using commutation of $d$ with restriction and $d^2=0$, \begin{align*} r_{k+2}(d\eta) &= (d(\eta|_U),d(\eta|_V)) \\ &= (d(d\alpha),d(d\beta)) \\ &= (0,0). \end{align*} Therefore $d\eta=0$, and $[\eta] \in H^{k+1}_{\mathrm{dR}}(M)$ is defined.[/step]

[guided]Start with a closed form $\gamma \in \Omega^k(U \cap V)$ and a lift $(\alpha,\beta) \in \Omega^k(U)\oplus \Omega^k(V)$ satisfying \begin{align*} \beta|_{U \cap V} - \alpha|_{U \cap V} = \gamma. \end{align*} The connecting map is supposed to differentiate this lift and then glue the result. Define \begin{align*} (a,b) := (d\alpha,d\beta) \in \Omega^{k+1}(U) \oplus \Omega^{k+1}(V). \end{align*} To glue $(a,b)$ to a global form on $M$, we must check that $(a,b)$ lies in $\ker s_{k+1}$. Since restriction commutes with exterior differentiation, \begin{align*} s_{k+1}(a,b) &= d\beta|_{U \cap V} - d\alpha|_{U \cap V} \\ &= d(\beta|_{U \cap V} - \alpha|_{U \cap V}) \\ &= d\gamma \\ &= 0. \end{align*} Thus $(a,b)\in \ker s_{k+1}$. Exactness of the Mayer-Vietoris sequence at $\Omega^{k+1}(U)\oplus\Omega^{k+1}(V)$ says precisely that every element of $\ker s_{k+1}$ is in the image of $r_{k+1}$. Hence there exists $\eta \in \Omega^{k+1}(M)$ such that \begin{align*} \eta|_U=d\alpha, \qquad \eta|_V=d\beta. \end{align*} It remains to check that $\eta$ defines a de Rham cohomology class, so we need $d\eta=0$. Restricting $d\eta$ to $U$ and $V$ gives \begin{align*} (d\eta)|_U=d(\eta|_U)=d(d\alpha)=0, \qquad (d\eta)|_V=d(\eta|_V)=d(d\beta)=0. \end{align*} Equivalently, \begin{align*} r_{k+2}(d\eta)=(0,0). \end{align*} Exactness at $\Omega^{k+2}(M)$ gives injectivity of $r_{k+2}$, so $d\eta=0$. Therefore $\eta$ is closed and $[\eta]\in H^{k+1}_{\mathrm{dR}}(M)$ is a legitimate output of the construction.[/guided]

[step:Show that changing the lift changes the global form by an exact form]Fix the closed representative $\gamma \in \Omega^k(U \cap V)$. Let $(\alpha,\beta)$ and $(\alpha',\beta')$ be two lifts in $\Omega^k(U)\oplus\Omega^k(V)$ with \begin{align*} s_k(\alpha,\beta)=\gamma, \qquad s_k(\alpha',\beta')=\gamma. \end{align*} Let $\eta,\eta' \in \Omega^{k+1}(M)$ be the corresponding global forms, so \begin{align*} r_{k+1}(\eta)=(d\alpha,d\beta), \qquad r_{k+1}(\eta')=(d\alpha',d\beta'). \end{align*} Define \begin{align*} (p,q):=(\alpha'-\alpha,\beta'-\beta) \in \Omega^k(U)\oplus\Omega^k(V). \end{align*} Then \begin{align*} s_k(p,q) &= s_k(\alpha',\beta')-s_k(\alpha,\beta) \\ &= \gamma-\gamma \\ &=0. \end{align*} By exactness at $\Omega^k(U)\oplus\Omega^k(V)$, there exists $\omega \in \Omega^k(M)$ such that \begin{align*} r_k(\omega)=(p,q). \end{align*} Using commutation of $d$ with restriction, \begin{align*} r_{k+1}(d\omega) &=(d(\omega|_U),d(\omega|_V)) \\ &=(d(\alpha'-\alpha),d(\beta'-\beta)) \\ &= r_{k+1}(\eta'-\eta). \end{align*} Since $r_{k+1}$ is injective, exactness at $\Omega^{k+1}(M)$ gives \begin{align*} \eta'-\eta=d\omega. \end{align*} Thus $[\eta']=[\eta]$ in $H^{k+1}_{\mathrm{dR}}(M)$.[/step]

[guided]Now keep the same closed representative $\gamma$, but allow two different lifts of it: \begin{align*} s_k(\alpha,\beta)=\gamma, \qquad s_k(\alpha',\beta')=\gamma. \end{align*} The possible ambiguity is that these two choices might produce different global forms $\eta$ and $\eta'$. We will show that the difference is exact on $M$. Define the difference of the two lifts by \begin{align*} (p,q):=(\alpha'-\alpha,\beta'-\beta) \in \Omega^k(U)\oplus\Omega^k(V). \end{align*} Because $s_k$ is linear, \begin{align*} s_k(p,q) &=s_k(\alpha'-\alpha,\beta'-\beta) \\ &=s_k(\alpha',\beta')-s_k(\alpha,\beta) \\ &=\gamma-\gamma \\ &=0. \end{align*} Thus $(p,q)\in \ker s_k$. Exactness at $\Omega^k(U)\oplus\Omega^k(V)$ identifies this kernel with $\operatorname{im} r_k$, so there exists a global $k$-form $\omega\in \Omega^k(M)$ satisfying \begin{align*} \omega|_U=\alpha'-\alpha, \qquad \omega|_V=\beta'-\beta. \end{align*} Differentiate this global form. Since exterior differentiation commutes with restriction, \begin{align*} (d\omega)|_U=d(\omega|_U)=d(\alpha'-\alpha)=d\alpha'-d\alpha, \end{align*} and similarly \begin{align*} (d\omega)|_V=d(\omega|_V)=d(\beta'-\beta)=d\beta'-d\beta. \end{align*} The global forms $\eta$ and $\eta'$ were defined by \begin{align*} \eta|_U=d\alpha,\quad \eta|_V=d\beta, \qquad \eta'|_U=d\alpha',\quad \eta'|_V=d\beta'. \end{align*} Therefore $d\omega$ and $\eta'-\eta$ have the same restrictions to $U$ and $V$: \begin{align*} r_{k+1}(d\omega)=r_{k+1}(\eta'-\eta). \end{align*} Exactness at $\Omega^{k+1}(M)$ gives injectivity of $r_{k+1}$, so \begin{align*} \eta'-\eta=d\omega. \end{align*} Hence $\eta'$ and $\eta$ differ by an exact form, and so they represent the same class in $H^{k+1}_{\mathrm{dR}}(M)$.[/guided]

[step:Show that changing the representative changes the global form by an exact form]Let $\gamma,\gamma' \in \Omega^k(U \cap V)$ be closed forms representing the same cohomology class in $H^k_{\mathrm{dR}}(U\cap V)$. Then there exists $\theta \in \Omega^{k-1}(U \cap V)$ such that \begin{align*} \gamma'=\gamma+d\theta, \end{align*} with the convention that for $k=0$ this means $\gamma'=\gamma$. For $k=0$, the assertion is exactly the lift-independence already proved. Assume $k\geq 1$. Choose $(\alpha,\beta)\in \Omega^k(U)\oplus\Omega^k(V)$ with $s_k(\alpha,\beta)=\gamma$. By surjectivity of $s_{k-1}$, choose $(\lambda,\mu)\in \Omega^{k-1}(U)\oplus\Omega^{k-1}(V)$ such that \begin{align*} s_{k-1}(\lambda,\mu)=\theta. \end{align*} Define \begin{align*} (\alpha',\beta') := (\alpha+d\lambda,\beta+d\mu) \in \Omega^k(U)\oplus\Omega^k(V). \end{align*} Then \begin{align*} s_k(\alpha',\beta') &=s_k(\alpha,\beta)+s_k(d\lambda,d\mu) \\ &=\gamma+d(s_{k-1}(\lambda,\mu)) \\ &=\gamma+d\theta \\ &=\gamma'. \end{align*} Let $\eta,\eta' \in \Omega^{k+1}(M)$ be the global forms produced from $(\gamma,\alpha,\beta)$ and $(\gamma',\alpha',\beta')$, respectively. Since $d^2=0$, \begin{align*} r_{k+1}(\eta') &=(d\alpha',d\beta') \\ &=(d\alpha+d^2\lambda,d\beta+d^2\mu) \\ &=(d\alpha,d\beta) \\ &=r_{k+1}(\eta). \end{align*} Injectivity of $r_{k+1}$ gives $\eta'=\eta$. By lift-independence, any other lift chosen for $\gamma'$ produces the same cohomology class. Hence the value of $\delta([\gamma])$ depends only on the cohomology class $[\gamma]$.[/step]

[guided]Now suppose the representative of the input class changes. Thus $\gamma$ and $\gamma'$ are closed $k$-forms on $U\cap V$ and represent the same de Rham cohomology class. This means that there is a form $\theta\in \Omega^{k-1}(U\cap V)$ such that \begin{align*} \gamma'=\gamma+d\theta. \end{align*} When $k=0$, there are no $(-1)$-forms, and cohomologous closed $0$-forms are equal; so this case has already been handled by the independence of the lift. We therefore assume $k\geq 1$. Choose a lift $(\alpha,\beta)\in \Omega^k(U)\oplus\Omega^k(V)$ of $\gamma$, so \begin{align*} s_k(\alpha,\beta)=\gamma. \end{align*} To compare with $\gamma'=\gamma+d\theta$, we lift the lower-degree form $\theta$. Surjectivity of \begin{align*} s_{k-1}: \Omega^{k-1}(U)\oplus\Omega^{k-1}(V)\to \Omega^{k-1}(U\cap V) \end{align*} gives forms $\lambda\in \Omega^{k-1}(U)$ and $\mu\in \Omega^{k-1}(V)$ such that \begin{align*} s_{k-1}(\lambda,\mu)=\theta. \end{align*} Now define a new lift candidate by adding the differentiated lower-degree lift: \begin{align*} (\alpha',\beta') := (\alpha+d\lambda,\beta+d\mu). \end{align*} This is a pair of $k$-forms on $U$ and $V$. We check that it lifts $\gamma'$: \begin{align*} s_k(\alpha',\beta') &=s_k(\alpha+d\lambda,\beta+d\mu) \\ &=s_k(\alpha,\beta)+s_k(d\lambda,d\mu) \\ &=\gamma+d(s_{k-1}(\lambda,\mu)) \\ &=\gamma+d\theta \\ &=\gamma'. \end{align*} The key point is that the correction term is a differential, so it disappears after applying $d$ once more. Let $\eta$ be the global form obtained from $(\alpha,\beta)$ and let $\eta'$ be the global form obtained from $(\alpha',\beta')$. Then \begin{align*} r_{k+1}(\eta') &=(d\alpha',d\beta') \\ &=(d(\alpha+d\lambda),d(\beta+d\mu)) \\ &=(d\alpha+d^2\lambda,d\beta+d^2\mu) \\ &=(d\alpha,d\beta) \\ &=r_{k+1}(\eta). \end{align*} Since $r_{k+1}$ is injective, $\eta'=\eta$. Thus replacing $\gamma$ by the cohomologous representative $\gamma'$ does not change the resulting cohomology class, at least for the specific lift constructed above. If a different lift of $\gamma'$ is chosen, the previous step shows that the resulting global form is cohomologous to $\eta'$. Therefore the value of $\delta$ depends only on the class $[\gamma]$.[/guided]

[step:Prove linearity of the induced map] Let $[\gamma_1],[\gamma_2]\in H^k_{\mathrm{dR}}(U\cap V)$ and let $c_1,c_2\in \mathbb{R}$. Choose closed representatives $\gamma_1,\gamma_2\in \Omega^k(U\cap V)$ and lifts \begin{align*} (\alpha_i,\beta_i)\in \Omega^k(U)\oplus\Omega^k(V), \qquad s_k(\alpha_i,\beta_i)=\gamma_i, \end{align*} for $i\in\{1,2\}$. Let $\eta_i\in \Omega^{k+1}(M)$ satisfy \begin{align*} r_{k+1}(\eta_i)=(d\alpha_i,d\beta_i). \end{align*} By linearity of $s_k$, \begin{align*} s_k(c_1\alpha_1+c_2\alpha_2,c_1\beta_1+c_2\beta_2) = c_1\gamma_1+c_2\gamma_2. \end{align*} The global form associated to this lift is $c_1\eta_1+c_2\eta_2$, because \begin{align*} r_{k+1}(c_1\eta_1+c_2\eta_2) &=c_1r_{k+1}(\eta_1)+c_2r_{k+1}(\eta_2) \\ &=(d(c_1\alpha_1+c_2\alpha_2),d(c_1\beta_1+c_2\beta_2)). \end{align*} Therefore \begin{align*} \delta(c_1[\gamma_1]+c_2[\gamma_2]) &=\delta([c_1\gamma_1+c_2\gamma_2]) \\ &=[c_1\eta_1+c_2\eta_2] \\ &=c_1[\eta_1]+c_2[\eta_2] \\ &=c_1\delta([\gamma_1])+c_2\delta([\gamma_2]). \end{align*} Thus $\delta$ is linear, and the preceding steps prove that it is well-defined. [/step]