I have some problems understanding a statement in the book by Warner "Foundations of Differential Manifolds and Lie Groups". This comes from section 2.33, page 75 but I will write a self-contained question.
Consider a smooth map $f:N \rightarrow M$ and some collection of 1-forms $\{\omega_i\}$ on $M$. Let $\pi_1,\pi_2 $ be the projections from $ N\times M $ to $N$ and $M$. For each $i$ define a form \begin{equation} \mu_i=\pi_1^* f^*\omega_i-\pi_2^*\omega_i. \end{equation} Let $\mathcal{I}$ be the ideal generated by $\{\mu_i\}$. Then it is proved in the book that the graph of $f$ is an integral manifold of $\mathcal{I}$, basically by noting that defining $g:N \rightarrow M$, $g(n)=(n,f(n))$ one has $g^*\omega_i=0$, hence the tangent space of the graph of $f$ is annihilated by $\mathcal{I}$.
The above seems to hold whatever collection of forms $\{\omega_i\}$ on $M$ one is considering. On the other hand, by Froebenius' theorem $\mathcal{I}$ admits an integral manifold if and only if it is a differential ideal. In order for $\mathcal{I}$ to be a differential ideal, that is $\mathrm{d}\mu_i\in \mathcal{I}$ for all $i$, it seems to me that some condition on the collection $\{\omega_i\}$ needs to be imposed - basically that $\mathrm{d}\omega_i\in\mathcal{I}$ which I do not think is automatic. Hence a contradiction.
Where am I going wrong?
The situation you described starts with the assumption that there is a map $f\colon N\to M$, and in that case the graph of $f$ is automatically an integral manifold of the ideal generated by the $\mu_i$'s. There's no need to assume integrability in that case, because the integral manifold is already given. That discussion was just to motivate the more substantive one to follow.
Farther down that same page, Warner shows how to use this idea to cook up a map when the map is not known in advance. In that case, he assumes that the $\omega_i$'s form a global coframe for $T^*M$, and he uses certain forms on $N$ to cook up an ideal on $N\times M$. Then -- and this is the crucial point -- he says "If $\mathscr I$ happens to be a differential ideal, then we can obtain the desired map $f$ (at least locally) from an integral manifold of $\mathscr I$." [Emphasis added.]