In this set of notes, Professor Terence Tao claims that there is a "duality" between forms and curves, where this claim is substantiated by the linearity of integration:
$$\int_\gamma(\omega_1+\omega_2)=\int_\gamma \omega_1+\int_\gamma\omega_2$$
and also that integration respects concatenation, in the sense that if we take $\gamma_1\cdot \gamma_2$ as a concatenation of paths, we can also see that $$\int_{\gamma_1\cdot\gamma_2} \omega=\int_{\gamma_1} \omega+\int_{\gamma_2} \omega$$ which generalizes the usual notion for intervals in single variable calculus.
I'm trying to understand this, since Tao mentions we can use homology. If we take
the integral along a path does not depend "choice of path," as in homotopic paths have the same integral.
In this case, there is a well defined homomorphism: if we take $[\gamma] \in H_1(M)$,
$$[\gamma] \mapsto\int_\gamma \omega $$
where concatenation is respected. Basically, my idea is that this is now a member of $\mathrm{Hom}(H_1(M),\mathbb R)\cong H^1(M)$.
In this way, we have an assignment $\Omega^1(M) \to H^1(M)$ (I actually don't know anything, I'm assuming it is trivially surjective for $M=\mathbb R^n$
The thing that I do not understand:
Question: How am I using the first equation here? And, is what I'm saying making sense?
remark: If someone could comment on how this observation relates to De Rham's theorem (if it does), I would be very happy.
First, there is one important thing you haven't quite stated right. The homotopy-invariance property is only true for closed $1$-forms (that is, $1$-forms $\omega$ such that $d\omega=0$). So you don't get a homomorphism $\Omega^1(M)\to H^1(M)$; you only get a homomorphism $Z^1(M)\to H^1(M)$ where $Z^1(M)$ is the space of closed $1$-forms. (Also, you should perhaps mention that $H^1(M)$ refers to cohomology with real coefficients, since usually it by default refers to integer coefficients.)
The first equation is then what tells you the map $Z^1(M)\to H^1(M)$ is a homomorphism. For it to be a homomorphism, you need to know that when you add differential forms, their integrals along any curve representing an element of $H_1(M)$ add as well, and that's what the first equation gives you.
The de Rham theorem says that for any manifold $M$, this map $Z^1(M)\to H^1(M)$ is surjective and its kernel is the set of exact $1$-forms: that is, the set of $1$-forms that are equal to $df$ for some smooth function $f$ on $M$. (Actually, the de Rham theorem says more than this; it says that a similar statement is valid for forms and cohomology of any degree, not just $1$.)