Identification of integration on smooth chains with ordinary integration

Solution 1:

We may assume that $A=[\sum\sigma_i]$ is actually just represented as the sum of all the $n$-simplices in some oriented smooth triangulation of $M$. Since the two integration maps differ by a constant multiple, it suffices to compare them on just a single $n$-form $\omega$ with nonzero integral. Pick $\omega$ to be supported on the interior of $\sigma_1$ such that $\int_{\Delta^n} \sigma_1^*(\omega)=1$. We then have $\int_{\Delta^n} \sigma_i^*(\omega)=0$ for $i\neq 1$, so your second integral sends $\omega$ to $1$.

We now compute $\int_M \omega$ as follows. Note that $\sigma_1$ itself is an oriented smooth chart of $M$, when restricted to the interior of $\Delta^n$. We can now construct a partition of unity on $M$ subordinate to a covering by oriented charts which has as one of its functions a bump function $f$ supported on the interior of $\sigma_1$ which is $1$ on the entire support of $\omega$. When we compute $\int_M\omega$ using this partition of unity, all the terms vanish except the one corresponding to $f$, since $f$ is the only one that does not vanish on the support of $\omega$. By definition, we then have $$\int_M\omega=\int_{\Delta^n}\sigma_i^*(f\omega)=\int_{\Delta^n}\sigma_i^*(\omega)=1.$$ Thus both integrals send $\omega$ to $1$, so they are equal.