The current differential is continuous and squares to zero

If $k=n$, then $dT=0$ by the convention in [F1], hence it is continuous. Suppose $k<n$. In a chart, [F2] expresses each coefficient of $d\eta$ as a finite signed sum of first partial derivatives of coefficients of $\eta$. Therefore, for every compact coordinate set $K$ and order $m$, the order-$m$ seminorm of $d\eta$ on $K$ is bounded by a finite constant times the order-$(m+1)$ seminorms of $\eta$ on the same set. Since $T$ is continuous, continuity at zero and linearity give finitely many seminorms $p_j$ and a constant $C$ with $\mathrm{\mid }T(\xi )\mathrm{\mid }\le C{\max }_j{p}_j(\xi )$: continuity supplies a neighborhood on which $\mathrm{\mid }T\mathrm{\mid }<1$, and scaling any nonzero $\xi$ into that neighborhood gives the bound. Applying it to $\xi =d\eta$ proves that $\eta \mapsto (-1{)}^{k+1}T(d\eta )$ is continuous. Thus $dT$ is a current.

If $k<n-1$, then for every $\zeta \in {\mathrm{\Omega }}^{n-k-2}(X)$, applying the definition twice gives $⟨d^{2}T,\zeta ⟩=(-1{)}^{k+2}⟨dT,d\zeta ⟩=(-1{)}^{k+2}(-1{)}^{k+1}⟨T,d^2\zeta ⟩=0$ by [F3]. If $k=n-1$, then $dT$ has degree $n$ and its differential is zero by the top-degree convention in [F1]. If $k=n$, then $dT=0$ already by [F1]. Hence $d^{2}T=0$ also at both endpoints.