do Carmo Riemannian Geometry Exercise 2.3: definition of $\nabla$ for an immersion The following is Exercise 3 of Chapter 2 of my Brazilian edition of do Carmo's Riemannian Geometry:

Let $f: M^n \to \overline M^{n + k}$ be an immersion from a differentiable manifold $M$ to a Riemannian manifold $\overline M$. Assume in $M$ the Riemannian metric induced by $f$:
$$
\langle u, v \rangle_p = \langle df_p(u), df_p(v) \rangle_{f(p)}.
$$
Let $p \in M$ and $U \subset M$ be a neighborhood of $p$ such that $f(U) \subset \overline M$ be a submanifold of $\overline M$. Let $X, Y$ be vector fields on $f(U)$ and extend then to vector fields $\overline X, \overline Y$ on an open subset of $\overline M$. Define
$$
(\nabla_X Y)(p) = \text{tangential component of } \overline \nabla_{\overline X} \overline Y(p), 
$$
where $\overline \nabla$ is the Riemannian connection of $\overline M$. Prove that $\nabla$ is the Riemannain connection of $M$.

My questions are:

*

*$X$ and $Y$ are vector fields on $f(U) \subset M$. Then $\nabla_X Y(p)$ does not make sense. Shouldn't it be $\nabla_X Y(f(p))$? Also, $\nabla$ is not a connection on $M$, but on $f(U)$, isn't it? So what does the problem want us to prove?

*What does "tangential component" mean?

 A: There is a bit of abusing notation here. The goal is to define $\nabla $, which is a connection on the tangent bundle of $M$. Thus the goal IS to define $\nabla _X Y(p)$, where $p\in M$ and $X, Y$ are local vector fields of $M$ around $p$. (I think it is very important to know that they are not defining something at $f(p)$: there might be $p\neq q$ so that $f(p) = f(q)$)
The way to do so is
(1) push-forward the local vector fields $X, Y$ to $\mathrm df(X), \mathrm df(Y)$ respectively, if $X, Y$ are on $U$, then $\mathrm df(X), \mathrm df(Y)$ are on $f(U)$ (they abuse notations here, identifying $X, Y$ with $\mathrm df(X), \mathrm df(Y)$)
(2) extend $\mathrm df(X), \mathrm df(Y)$ to a local vector fields $\overline X, \overline Y$ respectively on $\overline M$ around $f(p)$, and
(3) define $\nabla_X Y(p) := \text{tangential component of }\overline\nabla _{\overline X} \overline Y (f(p))$ (Note that it is $f(p)$ on the right hand side. I guess it is a typo) As suggested in the comment, the tangent space at $T_{f(p)} \overline M$ split into $df (T_pM)$ and $(df (T_pM))^\perp$, the orthogonal complement. The tangential component are taken with respect to this decomposition. Thus the more precise definition should be
$$\nabla_X Y(p) :=(\mathrm df)^{-1} \bigg(  \text{tangential component of }\overline\nabla _{\overline X} \overline Y (f(p))\bigg)$$
I suppose they went on to show that $\nabla$ is well defined, independent of the extension $\overline X, \overline Y$. Indeed $\nabla $ is the Levi-Civita connection on $M$ with respect to the pullback metric
$$\langle u, v \rangle_p := \langle df_p(u), df_p(v) \rangle_{f(p)}.$$
