In do Carmo's Riemannian Geometry book, on page 125, he has the following discussion:
Let $f:M^n \rightarrow \overline{M}^k$ be an immersion where $k \geq n$.
Then for each $p \in M$, there exists a neighborhood $U \subset M$ of $p$ such that $f(U) \subset \overline{M}$ is a submanifold of $\overline{M}$.
He also says that we identify $U$ with the image $f(U)$ and each vector $v \in T_qM$ with its image $df_q(v) \in T_{f(q)}\overline{M}$.
I am confused about two things here. Firstly, why does this neigbhorhood $U$ exists, and secondly, is $f:U \rightarrow f(U)$ an isometry? I find it odd to say that we are identifing $U$ with $f(U)$ if it is not an isometry.
The neighborhood $U$ exists because of the inverse function theorem. We're using the local diffeomorphism $f$ to identify $U$ with $f(U)$. There are no mentions of Riemannian metric, so why are you bringing in the notion of isometry?