I have written a solution to Problem 2-5 of Lee's Introduction to Riemannian Manifolds. I am self-learning and I would like to know if my solution is correct. The problem has 3 parts, (a), (b) and (c).
Definitions. Let $\pi: M^m \rightarrow N^n$ be a smooth submersion, and let $g$ be a Riemannian metric on $ M$. Let $g$ be a Riemannian metric on $M$. Given $p \in M$, the vertical tangent space at $p$ is defined to be $V_p := \ker(d\pi_p)$, and the horizontal tangent space at $p$ is $H_p := (V_p)^\perp$, where the orthogonal complement is taken with respect to the inner product $g_p$. A vector field $X$ on $M$ is said to be horizontal if $X_p \in H_p$ for every $p \in M$. Vertical vector fields are defined analogously.
Proposition (Part (a)). Let $X$ be a smooth vector field on $M$. Then there exists unique smooth vector fields $X^H$ and $X^V$ on $M$ such that $X = X^H + X^V$, and $X^H$ is horizontal and $X^V$ is vertical.
Proof. Fix $p \in M$. Then since $T_p M = H_p \oplus V_p$, there exist unique vectors $X_p^H \in H_p$ and $X_p^V \in V_p$ such that $X_p = X_p^H + X_p^V$. Define a rough vector field $X^H:M \rightarrow TM$ by $p \mapsto (p, X_p^H)$, and define $X^V$ analogously.
It remains to show that $X^H$ and $X^V$ are smooth. To this end, fix $p \in M$. By the Rank Theorem, there exist charts $(U, (x^i))$ and $(V,(y^j))$ such that $\pi(U) \subseteq V$ and the coordinate representation of $\pi$ is given by $$\widehat \pi: (x^1,\ldots,x^n,x^{n+1},\ldots,x^m) \mapsto (x^1,\ldots,x^n).$$ Let $\frac{\partial}{\partial x^i}$ and $\frac{\partial}{\partial y^j}$ denote the coordinate frames induced by $(U, (x^i))$ and $(V,(y^j))$, respectively. Fix $q \in U$. Then, with respect to the coordinate vectors at $q$ and $\pi(q)$, the differential $d\pi_q:T_qM \rightarrow T_{\pi(q)}N$ has matrix representation $[I_{n \times n} \quad 0_{(m-n) \times (m-n)}]$. This implies that $d\pi_q\left(\frac{\partial}{\partial x^{n+1}} \Big|_q\right) = \cdots = d\pi_q\left(\frac{\partial}{\partial x^{m}} \Big|_q\right) = 0.$
Since $d\pi_q$ is surjective, we know that $\dim(\ker(d\pi_q)) = \dim(T_q M) - \dim(T_{\pi(q)}N) = m-n$. This implies that $$V_q = \ker(d\pi_q) = \text{span}\left(\frac{\partial}{\partial x^{n+1}}\Big|_q, \ldots, \frac{\partial}{\partial x^{m}}\Big|_q \right).$$
Next, by applying the Gram-Schmidt algorithm to the local frame $\left(\frac{\partial}{\partial x^{n+1}},\ldots, \frac{\partial}{\partial x^{m}},\frac{\partial}{\partial x^{1}},\ldots,\frac{\partial}{\partial x^{n}}\right)$ (the order is very important here), we obtain a smooth orthonormal frame $(E_1,\ldots,E_{m-n},E_{m-n+1},\ldots,E_m)$ on $U$ such that for each $q \in U$,
$$V_q = \text{span}\left(\frac{\partial}{\partial x^{n+1}}\Big|_q, \ldots, \frac{\partial}{\partial x^{m}}\Big|_q \right) = \text{span}\left(E_1|_q,\ldots,E_{m-n}|_q\right).$$
Orthogonality implies that $H_q = \text{span}(E_{m-n+1}|_q,\ldots, E_m|_q)$. Now, write $X = X^i E_i$. Then each $X^i:U \rightarrow \mathbb R$ is smooth, because $X^i = g(X,E_i)$. Finally, observe that on $U$, we have $$X^V = X^1 E_1 + \ldots + X^{m-n} E_{m-n}, \quad X^H = X^{m-n+1} E_{m-n+1} + \ldots + X^{m} E_{m}.$$ Thus, $X^H$ and $X^V$ are smooth, because they are smooth on a neighbourhood around every point in $M$. $\square$
Proposition (Part (b)). Let $Y$ be a smooth vector field on $N$. Then there exists a unique smooth horizontal vector field $X$ on $M$ such that $d\pi_p(X_p) = Y_{\pi(p)}$ for every $p \in M$.
We say that $X$ is the horizontal lift of $Y$.
Proof. Fix $p \in M$. Again, by the Rank Theorem, there exist charts $(U_p, (x^i))$ and $(V_p,(y^j))$ such that $\pi(U_p) \subseteq V_p$ and the coordinate representation of $\pi$ is given by $(x^1,\ldots,x^n,x^{n+1},\ldots,x^m) \mapsto (x^1,\ldots,x^n).$ Fix $q \in U_p$. The proof of Part (a) tells us that the matrix representation of $d\pi_q$ is $[I_{n \times n} \quad 0_{(m-n) \times (m-n)}]$. This implies that $d\pi_q\left(\frac{\partial}{\partial x^i}|_q\right) = \frac{\partial}{\partial y^i}|_{\pi(q)}$ for all $i = 1,\ldots,n$.
Write $Y = Y^j \frac{\partial}{\partial y^j}$. Consider the smooth vector field $\sum_{i=1}^n (Y^i \circ \pi) \frac{\partial}{\partial x^i}$. Although this vector field is not necessarily horizontal, Part (a) implies that $X^{(p)} := \left(\sum_{i=1}^n (Y^i \circ \pi) \frac{\partial}{\partial x^i} \right)^H$ is a well-defined smooth horizontal vector field. Observe that for $q \in U_p$, $$d\pi_q(X^{(p)}|_q) = \sum_{i=1}^n Y^i|_{\pi(q)} d\pi_q\left(\frac{\partial}{\partial x^i}\Big|_q \right) = \sum_{i=1}^n Y^i|_{\pi(q)}\frac{\partial}{\partial y^i}\Big|_{\pi(q)} = Y_{\pi(q)}.$$
Next, let $\{\psi_p\}_{p \in M}$ be a smooth partition of unity subordinate to $\\{U_p\\}\_{p\in M}$. Define $X := \sum_{p \in M} \psi_p X^{(p)}$. Fix $q \in M$. Then $$d\pi_q(X_q) = \sum_{p \in M} \psi_p(q) d\pi_q(X^{(p)}|_q) = \sum_{p \in M} \psi_p(q) Y_{\pi(q)} = Y_{\pi(q)}.$$
Finally, to show uniqueness, suppose $Z$ is another horizontal smooth vector field on $M$ such that $d\pi_q(Z_q) = Y_{\pi(q)}$ for all $q \in M$. Observe that the restriction $d\pi_q|_{H_q}:H_q \rightarrow T_{\pi(q)} N$ is injective, so $X_q = Z_q$. $\square$
Proposition (Part (c)). Fix $p \in M$ and $v \in H_p$. Then there exists a vector field $Y$ on $N$ such that its horizontal lift $X$ satisfies $X_p = v$.
Proof. Let $Y$ be any vector field on $N$ such that $Y_{\pi(p)} = d\pi_p(v)$. Let $X$ denote the horizontal lift of $Y$. Then $d\pi_p(X_p) = Y_{\pi(p)} = d\pi_p(v)$. Since $d\pi_q|_{H_q}:H_q \rightarrow T_{\pi(q)} N$ is injective, we have $X_p = v$, as we wished. $\square$