A compact oriented source cycle defines a closed current

By [F6], cover $\mathrm{\Sigma }$ by finitely many ordinary slice charts and localize the integral with finitely many smooth cutoffs. In such a chart, $i$ is the coordinate inclusion $u\mapsto (u,0)$, so the pointwise pullback formula [F2] makes each coefficient of $i^{\ast }\eta$ a restriction of a coefficient of $\eta$ (with the fixed coordinate inclusion differential). These coefficients are bounded on the compact localized supports. Each chart integral is therefore bounded by a constant times an order-zero $C^{\mathrm{\infty }}$ seminorm of $\eta$ on a compact subset of $X$. Summing the finitely many bounds gives $\mathrm{\mid }{\delta }_{\mathrm{\Sigma }}(\eta )\mathrm{\mid }\le C{p}_0(\eta )$ for a continuous seminorm $p_0$ on ${\mathrm{\Omega }}^p(X)$. Thus ${\delta }_{\mathrm{\Sigma }}$ is continuous and is a current.

Use a slice chart from [F6], writing $x=(u^1,\dots ,u^p,v^1,\dots ,v^{n-p})$ and $i(u)=(u,0)$. Write an ambient $r$-form as $\eta ={\sum }_I{a}_I(u,v)d{x}^I$ with increasing multi-indices $I$. By [F2] and [F3], any wedge containing a $d{v}^j$ pulls back to zero, while a purely tangential wedge $d{u}^I$ pulls back to itself. Therefore [F4] gives $d(i^{\ast }\eta )={\sum }_{I\subseteq \{1,\dots ,p\},j\le p}({\mathrm{\partial }}_{u^j}{a}_I)(u,0)d{u}^j\wedge d{u}^I=i^{\ast }(d\eta )$: differentiating $a_I(u,0)$ in a tangential coordinate is precisely the corresponding partial derivative of $a_I$ at $(u,0)$, and every term of $d\eta$ containing a normal differential vanishes on pullback. This calculation includes $r=0$, empty sums for $r>p$, and $p=0$, where all positive-degree forms on $\mathrm{\Sigma }$ vanish. Equality in a slice chart at every point proves the global identity.

If $p=0$, then ${\delta }_{\mathrm{\Sigma }}$ has degree $n$ and its differential is zero by the top-degree convention in [F1]. If $p>0$, take any ambient test form $\eta \in {\mathrm{\Omega }}^{p-1}(X)$. By the differential definition and step 1.2, $⟨d{\delta }_{\mathrm{\Sigma }},\eta ⟩=(-1{)}^{n-p+1}{\int }_{\mathrm{\Sigma }}{i}^{\ast }(d\eta )=(-1{)}^{n-p+1}{\int }_{\mathrm{\Sigma }}d(i^{\ast }\eta )=0$ by [F5]. The test form is ambient, as required by the current domain. Hence $d{\delta }_{\mathrm{\Sigma }}=0$ in all dimensions.