Smooth forms define currents compatibly with the differential

Fix a finite chart localization of $X$. On each chart, the coefficients of the fixed smooth form $\alpha$ are bounded on the compact support of the localized top-form integral. The coefficient of $\alpha \wedge \eta$ is a finite sum of products of such coefficients with coefficients of $\eta$. The Euclidean integral of each localized term is therefore bounded by a constant times an order-zero seminorm of $\eta$ on the finite chart cover. Summing the finitely many bounds proves that $T_{\alpha }$ is continuous; its linearity follows from linearity of wedge and chart integration.

If $k=n$, both $d{T}_{\alpha }$ and $T_{d\alpha }$ are zero by the out-of-range current convention and the absence of $(n+1)$-forms. If $k<n$, take $\eta \in {\mathrm{\Omega }}^{n-k-1}(X)$. By [F2] and [F3], $0={\int }_{X}d(\alpha \wedge \eta )={\int }_{X}d\alpha \wedge \eta +(-1{)}^k{\int }_X\alpha \wedge d\eta \mathrm{.}$ Hence $(-1{)}^{k+1}{\int }_X\alpha \wedge d\eta ={\int }_{X}d\alpha \wedge \eta$. By [F1] the left side is $⟨d{T}_{\alpha },\eta ⟩$ and the right side is $⟨T_{d\alpha },\eta ⟩$. Equality for every test form proves $d{T}_{\alpha }=T_{d\alpha }$.