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Link to paper; proofreading
LSpice
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How to see $\delta_2(\hat{\chi}(V))=\chi(V)$ in differential cohomology?

I'm reading the paper "Differential Characters and Geometric Invariants" by Cheeger and Simons. In Page 62 the authors defined the differential Euler character $\hat{\chi}(V)\in \hat{H}^{2n-1}(M, \mathbb{R}/\mathbb{Z})$ for an oriented $2n$-dimensional bundle $\pi:V\to M$ as follows: Let $\theta$ be a connection on $V$ with curvature $\Omega$. Let $P_{\chi}$ be the Pfaffian. Let $SV$ be the sphere bundle associated with $V$ and $Q$ be the canonical $2n-1$-form on $SV$ such that $$ \pi^*(P_{\chi}(\Omega))=dQ \text{, and } \int_{S^{2n-1}}Q=1 \text{ on each fiber.} $$ For a cycle $z\in Z_{2n-1}(M)$, we can find a cycle $y\in Z_{2n-1}(SV)$ and $w\in C_{2n}(M)$ such that $$ z=\pi_*(y)+\partial w. $$ Then $\hat{\chi}(V): Z_{2n-1}(M)\to \mathbb{R}/\mathbb{Z}$ is defined to be $$ \hat{\chi}(V)(z)=\widetilde{Q(y)}+\widetilde{P_{\chi}(\Omega)(w)}, $$ where $\widetilde{(\cdot)}$ denotes the modulo $\mathbb{Z}$ map.

Now I want to compute the image of $\hat{\chi}(V)$ under the map $\delta_2: \hat{H}^{2n-1}(M, \mathbb{R}/\mathbb{Z})\to H^{2n}(M, \mathbb{Z})$. According to the construction in Page 53, we can first find a real-valued cochain $T: C^{2n-1}(M,\mathbb{R})$ such that $$ \widetilde{T}\rvert_{Z_{2n-1}(M)}=\hat{\chi}(V). $$ Then we can show that we can find a differential form $\alpha\in \wedge^{2n}(M)$ and an integer-valued cochain $c\in C^{2n}(M,\mathbb{Z})$ such that $$ \delta T=\alpha-c. $$ We can further show that $\alpha$ and $c$ are both closed and we define $$ \delta_2 (\hat{\chi}(V)):=[c]\in H^{2n}(M,\mathbb{Z}). $$ In page 62 the author said it is not difficult to show that $\delta_2 (\hat{\chi}(V))=\chi(V)$, the integral Euler class. But it is not that obvious to me.

My question is: for the so constructed $\hat{\chi}(V)$, how to find a real-valued cochain $T: C^{2n-1}(M,\mathbb{R})$ such that $$ \widetilde{T}\rvert_{Z_{2n-1}(M)}=\hat{\chi}(V)? $$ And then how to compute $\delta_2 (\hat{\chi}(V))$?

Zhaoting Wei
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