Vrftsjtryk

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 $intlimits_{partial_i s} eta$ just means $intlimits_{Delta_{p-1}} (s circ F_i)^{ast} eta$, 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 $deta$, and therefore

$$intlimits_s deta = intlimits_{Delta_p} d s^{ast} eta$$

$$intlimits_{partial s} eta = sumlimits_i (-1)^i intlimits_{Delta_{p-1}} F_i^{ast}( s^{ast} eta)$$

So we can replace $eta$ by its pullback $s^{ast}eta$, and forget about $M$ entirely.