Let $D$ be a distribution. Show that every point of $p \in M$ has a submanifold $N$ such that $p \in M$ and $T_p N = D_p$. Let $D$ be a distribution. Show that every point of $p \in M$ has an embedded submanifold $N$ such that $p \in M$ and $T_p N = D_p$.
Attempt: Assume $M$ is a smooth manifold of dimension $m$ and $D$ is a rank $k$-subbundle of $\pi: TM \to M$. 
Let $p \in M$. Somehow, we will have to construct an embedded submanifold $N$ of dimension $k$ such that $D_p = T_p N$. 
How can I construct such an embedded manifold? Any hints will be appreciated!
 A: The problem is to find a submanifold that takes a certain tangent space at a certain point. You can do this when your manifold is an open subset of $\mathbb{R}^n$ trivially; just take a little piece of an affine space. Luckily, every manifold is locally diffeomorphic to an open subset of $\mathbb{R}^n$.
More specifically: let $\varphi\colon U \to \mathbb{R}^n$ be a chart of the manifold at $p$. Let $N$ be an embedded submanifold in $\varphi(U) \subseteq \mathbb{R}^n$ that takes the tangent space $(d \varphi)_p(D_p)$ at $\varphi(p)$. (For example, let $N$ be a restriction of an affine subspace.) Then, $\varphi^{-1}(N)$ is an embedded submanifold of $U$, and hence of $M$, that takes the tangent space $D_p$ at $p$. (To verify this, it is most convenient to regard an embedded submanifold as an inclusion map, and the tangent space "taken at a point" as the image of the differential of this inclusion.)
A: One can proceed as Christopher Gadzinski indicated in his answer. Here is another approach.
Let $g$ be a Riemannian metric on $M$. For each $v \in T_pM$ there is a unique geodesic $\gamma_v$ with $\gamma_v'(0) = v$. There is a connected open subset $U \subseteq T_pM$ with $0 \in U$, such that if $v \in U$, then the geodesic $\gamma_v$ is defined at $1$. The exponential map at $p$, denoted $\operatorname{exp}_p : U \to M$, is given by $\operatorname{exp}_p(v) = \gamma_v(1)$. If $r$ denotes the injectivity radius at $p$, then $\operatorname{exp}_p : B(0, r) \to M$ is a diffeomorphism onto its image. As $B(0, r)\cap D_p$ is an embedded submanifold of $B(0, r)$ of dimension $k$, then $N := \operatorname{exp}_p(B(0, r)\cap D_p)$ is an embedded submanifold of $M$ of dimension $k$.
It now follows that $T_pN = D_p$. To see this, first note that for $t \in \mathbb{R}$ such that $tv \in U$, we have $\gamma_v(t) = \operatorname{exp}_p(tv)$. In particular, for every $v \in D_p$, there is $\varepsilon > 0$ such that $\gamma_v((-\varepsilon, \varepsilon)) \subseteq N$. Therefore $v = \gamma_v'(0) \in T_pN$, so $D_p \subseteq T_pN$, but $\dim D_p = k = \dim T_pN$ so they are equal.
This idea is often used to demonstrate the relationship between sectional curvature and Gaussian curvature: if $P \subseteq T_pM$ is a two-dimensional subspace, then $\operatorname{sec}_p(P) = K_p(N)$.
A: This is not always true there exist  distributions which are not integrable.
https://mathoverflow.net/questions/294276/a-strongly-non-integrable-distribution
