Let $n \geq m$. Is it true that for any immersion $\theta: M^m \to N^n$ and $m$-form on $N$, we have $$ \int_{M} \theta^\ast \alpha = \int_{\theta(M)} \alpha? $$
Is is true for submersions? Or even for arbitrary smooth maps? Although very elementary, I keep confusing myself. (Do we have to argue through 'immersions are locally embeddings, use partition of unity, diffeomorphism invariance of the integral and blabla'?)
The question arose in the following setting: Take the immersion $\gamma: S^1 \to \mathbb{R}^2$ via $(x,y) \mapsto (x,xy)$, which is the figure 8. I want to show that this figure encloses a region $Z$ of oriented area 0.
So,
$$ \int_{Z} dp \wedge dq = \int_{\partial Z} p dq = \int_{\gamma(S^1)} p dq \overset{?}= \int_{S^1}\gamma^\ast(p dq) = 0, $$ where the last equality is an explicit computation in polar coordinates, which I omitted.
Edit: Perhaps, this is not true in general, but rather we have the special situation here, that $\gamma\vert_{S^1 - \{-1\}}$ actually has the same trace as $\gamma$. Moreover, it is injective, and thus an embedding (since $S^1$ is compact). And then, clearly, the diffeomorphism invariance kicks in. (The removal of a point does not matter for the integral).
With this, I shall conclude that: The above equality holds for any immersion which is an embedding away from a set of measure zero.
Consider $M = N = S^1$, and $\theta : S^1 \to S^1 : u \mapsto 2u \bmod 2 \pi$. Note that $\theta$ is an immersion, as required, and a submersion as well.
Now $\theta(S^1)$ is (as a set) all of $S^1$, albeit double-covered.
If $\alpha$ is any form for which $\int_{S^1} \alpha = C \ne 0$, then \begin{align} \int_{M} \theta^\ast \alpha &= \int_{S^1} \theta^\ast \alpha \\ &= 2 \int_{S^1} \alpha \\ &= 2C \end{align} while \begin{align} \int_{\theta(M)} \alpha & = \int_{S^1} \alpha \\ &= C \\ \end{align}
And because $C$ is nonzero, $C$ and $2C$ are different.
In short: your proposed equality is not true in general.
"But what if the immersion is 'nice', like a figure-8?" you might ask. What if the set of double-points in the image (perhaps "double values") has measure zero?"
In that case, you could let $B$ be the set of points $p \in M$ such that there's another point $q\in M$ with $\theta(p) = \theta(q)$, i.e., the set of "multiple points" in the domain, and let $N = M - B$. Something like Sard's theorem, I think, ensures that $B$, too, has measure zero. And then you have \begin{align} \int_{M} \theta^\ast \alpha &= \int_{N} \theta^\ast \alpha \\ &= \int_{\theta(N)} \alpha & \text{by the chain rule}\\ &= \int_{\theta(M)} \alpha & \text{because $\theta(B)$ also has measure zero}. \end{align}
I've been pretty sloppy here, but I think all the essentials are there. I suspect all the theorems you need are in Milnor's "Topology from the Differentiable Viewpoint", for instance, combined with some book that talks carefully about the chain rule for $k$-forms.