Let $f,g: M^{k} \to N$ ($M$ and $N$ with out boundary ) such that they are homotopic then for $\omega$ a $k$-form on $N$ do we have that
$$ \int_M f^{\ast} \omega = \int_M g^{\ast} \omega$$
as conclusion? I can't figure out a proof so I am starting to think that it is not true. I can't use Homology or Cohomology and all that fancy stuff . I think is just a matter of rearranging singular cubes but I can't see how.
The way I was trying to approach this is the following: Consider first the case where support of $\omega$ is in the interior of the image of a singular cube $f \circ c$, where $c$ is a singular cube into $M$.
Then we look at the new singular cube $f \circ c : I \to N$. But I don't know what to do from here and how to approach the general case.
Attempt
Using @TedShifrin's answer I've managed the following
$$\int_{M \times [0,1]}d(H^{*}w)=\int_{\partial(M \times [0,1])}H^{*}w = \int_{\partial M}g^{\ast}w-f^{\ast}w=0$$
but I am not sure.
Thanks a lot in advance.
Take $M=N=\mathbb R$ and take $\omega$ to be some nonvanishing $1$-form. Any two functions $\mathbb R\to\mathbb R$ are homotopic to each other, hence all are homotopic to zero. If your claim were true it would imply that the integral of any $1$-form is zero, which of course is not the case!
What you might be thinking of is that if $f$ and $g$ are homotopic then they induce the same map on de Rham complexes. In the previous example, I showed that every map $\mathbb R\to\mathbb R$ induces zero as a function from the de Rham complex of $\mathbb R$ to itself. But because $\mathbb R$ is convex (or better, contractible) its de Rham cohomology vanishes! So we don't learn anything.