# Why is the transgression of differential forms a form?

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?

• Every scalar function is a $0$-form, so maybe that is all they mean. Dec 26, 2022 at 13:34
• the $\bullet$ in $\Omega^{\bullet}(U)$ indicates the dimension Dec 26, 2022 at 13:36

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