Why is the transgression of differential forms a form?

Question

In the nLab page on transgression of differential forms at definition 2.7 they have

Let $E \stackrel{\mathrm{fb}}{\rightarrow} \Sigma$ be a field bundle over a spacetime $\Sigma$ (def. 2.5), with induced jet bundle $J_{\Sigma}^{\infty}(E)$. For $\Sigma_r \hookrightarrow \Sigma$ be a submanifold of spacetime of dimension $r \in \mathbb{N}$, then transgression of variational differential forms to $\Sigma_r$ is the function $$ \tau_{\Sigma_r}: \Omega_{\Sigma, \mathrm{cp}}^{r, \bullet}(E) \longrightarrow \Omega^{\bullet}\left(\Gamma_{\Sigma_r}(E)\right) $$ which sends a differential form $A \in \Omega_{\Sigma, \text { cp }}^{r, \bullet}(E)$ to the differential form $\tau_{\Sigma_r} \in \Omega^{\bullet}\left(\Gamma_{\Sigma_r}(E)\right)$ which to a smooth family on field configurations $$ \Phi_{(-)}: U \times N_{\Sigma} \Sigma_r \longrightarrow E $$ assigns the differential form given by first forming the pullback of differential forms along the family of jet prolongation $j_{\Sigma}^{\infty}\left(\Phi_{(-)}\right)$followed by the integration of differential forms over $\Sigma_r$ : $$ \tau_{\Sigma}A_{\Phi_{(-)}}:=\int_{\Sigma_r}\left(j_{\Sigma}^{\infty}\left(\Phi_{(-)}\right)\right)^*A \in \Omega^{\bullet}(U) . $$

Why is the last expression a differential form? After integration we have a number so isn't it a function?

Answer 1

After integration we have a number so isn't it a function?

Differential $k$-forms can be integrated along a submersion with $d$-dimensional fibers, which yields a differential $(k-d)$-form.

Fiberwise integration (alias pushforward) of differential forms is a standard operation, described in many expository texts. A fairly detailed presentation is given in Chapter VII of

• Werner Greub, Stephen Halperin, Ray Vanstone. Connections, Curvature, and Cohomology. Volume I. De Rham Cohomology of Manifolds and Vector Bundles. Pure and Applied Mathematics 47A (1972), Academic Press.