I am reading a proof of Stokes' theorem in Differentiable Manifolds by Conlon. Here $M$ is an $n$-dimensional smooth manifold, and $\Delta_p$ is the usual $p \leq n $-simplex in $\mathbb R^p$, and $s$ is a smooth map from $\Delta_p \rightarrow M$.
The boundary $\partial_i$ is a certain identification $F_i$ of $\Delta_{p-1}$ with one of the faces of $\Delta_p$. The integral $\int\limits_{\partial_i s} \eta$ just means $\int\limits_{\Delta_{p-1}} (s \circ F_i)^{\ast} \eta$, where the $^\ast$ means the pullback of $\eta$ to a top form on the interior of $\Delta_{p-1}$.
I am confused about the very beginning of the proof, "It is clearly sufficient to prove that...". There is no mention of $s$ anymore. In fact, it seems like the $n$ dimensional manifold $M$ is replaced by a $p$-dimensional manifold. This manifold is taken to be an open set $U$ in $\mathbb R^p$ containing $\Delta_p$, and it looks like $s$ is just taken to be the inclusion map from $\Delta_p$.
My question is: how we can reduce to this case?
I was thinking we could maybe find a $p$-dimensional submanifold $N$ of $M$ through which $s$ factors, and then use a partition of unity on $N$. This might at least reduce us to the case where $s$ is a smooth map $\Delta_p \rightarrow U \subset \mathbb R^k$, not necessarily the inclusion map.
I do not know whether there is always an embedded submanifold $N$ of $M$ which contains the image of $s$ though (I asked this in a previous question).
Okay actually this is not so complicated. The chain $s: \Delta_p \rightarrow M$ extends to a smooth function on an open neighborhood $U \subset \mathbb R^p$ containing the $p$-simplex. Then the pullback $s^{\ast} \eta$ is a differential $p-1$ form on $U$. The exterior derivative of $s^{\ast} \eta$ is the pullback of $d\eta$, and therefore
$$\int\limits_s d\eta = \int\limits_{\Delta_p} d s^{\ast} \eta$$
$$\int\limits_{\partial s} \eta = \sum\limits_i (-1)^i \int\limits_{\Delta_{p-1}} F_i^{\ast}( s^{\ast} \eta)$$
So we can replace $\eta$ by its pullback $s^{\ast}\eta$, and forget about $M$ entirely.