I would like to know how to prove that the pullback of a foliation is actually a well-defined concept.
To be precise, if $M$ is an $n$-dimensional smooth manifold, a $k$-dimensional foliation $\mathcal F$ of $M$ is a set of nonempty, connected, mutally disjoint, immersed $k$-dimensional submanifolds of $M$ called the leaves of the foliation, such that their union covers $M$ and for each $p\in M$ there exists a flat chart for $\mathcal F$, that is, a chart $(U,\varphi)$ of $M$ such that $\varphi(U)$ is a cube in $\mathbb R^n$ and the intersection of $U$ with each element of $\mathcal F$ is either the empty set or a countable union of slices of the form $x^{k+1}=c^{k+1},\dots,x^n=c^n$.
Lets recall that if $\mathcal F$ is a foliation of a smooth manifold $M$ and $f:N\rightarrow M$ is a smooth map transversal to the leaves of the foliation, then there exists a unique foliation on $N$, the pullback of $\mathcal F$, denoted by $f^*\mathcal F$, such that $f$ becomes a foliation-preserving map (it maps leaves of $f^*\mathcal F$ into leaves of $\mathcal F$).
I have seen how to apply the transversality of $f$ in order to construct a flat chart for $f^*\mathcal F$, but I don't know how to prove that the connected components of $f^{-1}(L)$, where $L$ is a leaf of $\mathcal F$, are actually immersed submanifolds of $N$. For instance, I don't know hot to prove that such sets are Hausdorff, etc., and I have not seen a proof of this fact anywhere. It may be a technicality, but proving that there are flat charts does not suffice to conclude that the given set is a foliation.
Thanks in advance for your answers.
Best Answer
What's needed to make this proof work well is a definition of foliation that is distinct from but equivalent to the definition that you have stated.
Starting from your definition, let me first broaden your notion of a foliation chart slightly: instead of the domain $U$ being a cube in $\mathbb R^n$, let the domain by a subset of $\mathbb R^n$ of the form $$U = U^\parallel \times U^\perp $$ where $U^\parallel \subset \mathbb R^k$ is open and $U^\perp \subset \mathbb R^{n-k}$ is open, and we have of course identified $\mathbb R^n = \mathbb R^k \times \mathbb R^{n-k}$. Also we require, of course, that $V = \phi(U) \subset M$ be open and that $\phi : U \to V$ be a diffeomorphism. Instead of the notation $(U,\phi)$ I'll just denote this foliation chart as $\phi : U \to V$.
So, your definition of a foliation produces a certain collection of foliation charts that cover $M$, and that have a "compatibility condition" with respect to the given collection of leaves.
By applying your definition of foliation, and then refining the given collection of charts, you can obtain a new collection of foliation charts that satisfies a more "localized" compatibility condition which is expressed solely as a condition on the overlap maps of the charts.
Namely, there exists a collection of foliation charts $\{\phi_i : U_i \to V_i\}_{i \in I}$ covering $M$ such that for any $i,j \in I$ the overlap map $$h_{ji} = \phi_j^{-1} \circ \phi_i : \phi_i^{-1}(V_i \cap V_j) \to \phi_j^{-1}(V_i \cap V_j) $$ can be written in the following form for all $(x,y) \in \phi_i^{-1}(V_i \cap V_j) \subset U_i^\parallel \times U_i^\perp$: $$h_{ji}(x,y) = (f_{ji}(x,y),g_{ji}(y)) \in U_j^\parallel \times U_j^\perp $$ where the functions $f_{ji} : U_i^\parallel \times U_i^\perp \to U_j^\parallel$ and $g_{ji} : U_i^\perp \to U_j^\perp$ are smooth. You may have to stare at this for a long time (I have had to do so repeatedly over the years), but the intuition is this. First, any value $\eta \in U_i^\perp$ determines a horizontal set $U_i^\parallel \times \{\eta\}$ --- whose image under $\phi_i$ is sometimes called a plaque or a local leaf. Furthermore, two points on the same local leaf in the domain of the overlap map must be taken to two points on the same local leaf in the range of the overlap map.
Once you have a collection of foliation charts in $M$ that satisfy this overlap condition, you can then deduce, using the implicit function theorem, that $N$ also has a collection of charts that satisfy this overlap condition, and furthermore the restriction of the map $f$ to each $N$-chart maps into some $M$-chart satisfying a condition that is very similar to the overlap condition.
So, the final step is prove a general statement, which amounts to finishing off the proof that the two definitions of foliation on $N$ are equivalent, namely your definition with foliation charts and leaves, and my definition with foliation charts and overlap maps. What's left to do in proving this equivalence is to assume that one has foliation charts and overlap maps, then to define the leaves, and finally to check the remaining portions of your definition. Let me do just the fun part which is defining the leaves. And I'll express this in $M$ (even though for your question you want to apply it in $N$).
So we start from an atlas of foliation charts $\{\phi_i : U_i \to V_i\}_{i \in I}$ in $M$ satisfying the overlap condition. To define leaves there are two steps, and then there's one last step.
In Step 1 you define the leaf topology on $M$, which is the topology generated by open subsets of plaques, i.e. by subsets the form $\phi_i(W \times \{y\})$ where $i \in I$ varies, $W \subset U_i^\parallel$ varies over open subsets of $U_i$, and $y \in U_i^\perp$ varies. You prove that the leaf topology gives $M$ the structure of a topological $k$-manifold, and by using those subsets and the corresponding maps $W \mapsto \phi_i(W \times \{y\})$ you get a smooth $k$-manifold structure on $M$. (It might be better to talk about the leaf atlas on $M$ than just the leaf topology, but the "leaf topology" terminology is fairly common).
In Step 2 you define the leaves of the foliation to be the connected components of the leaf topology on $M$; since $M$ is a smooth $k$-manifold in the leaf topology, each leaf is therefore also a smooth and connected $k$-manifold.
The last step is to prove that the inclusion map of each leaf is an smooth immersion, and to verify the compatibility clause of your definition of a foliation.