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Small typos fixed
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Signed-off-by: Marcello Seri <[email protected]>
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mseri committed Dec 5, 2023
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Expand Up @@ -103,15 +103,15 @@ \section{Inverse function theorem}
our choice of coordinates after the above observations implies that
$\det \left( \frac{\partial Q^i}{\partial x^j} \right) \neq 0$ at $(x,y) = (0,0)$.

Since the gradient of $Q$ with respect to $z$ is regular, we are going to extend
Since the gradient of $Q$ with respect to $x$ is regular, we are going to extend
the mapping with the identity on the rest of the coordinates to get a regular map
on the whole neighbourhood.
Let $\varphi : U \to \R^m$ be defined by $\varphi(x,y) = (Q(x,y), y)$. Then,
\begin{equation}
D\varphi(0,0) =
\begin{pmatrix}
\frac{\partial Q^i}{\partial x^j}(0,0) & \frac{\partial Q^i}{\partial y^j}(0,0) \\
0 & \id_{\R^k}
0 & \id_{\R^{m-k}}
\end{pmatrix}
\end{equation}
has nonvanishing determinant by hypothesis.
Expand Down Expand Up @@ -185,6 +185,12 @@ \section{Inverse function theorem}
concluding the proof.
\end{proof}

\begin{exercise}
Formulate and prove a version of the Rank theorem for a map $F : M^m \to N^n$ of constant rank $k$,
where $M$ is a smooth manifold with boundary, $N$ is a smooth manifold without boundary
and $\ker dF_p \not\subseteq T_p\partial M$.
\end{exercise}

\section{Embeddings, submersions and immersions}

Looking at the statement of the Rank Theorem, one can already see that there can be different possibilities depending on the relation between, $m$, $n$ and $k$. This warrants a definition.
Expand Down Expand Up @@ -261,7 +267,7 @@ \section{Embeddings, submersions and immersions}
In the rest of this chapter we will try to give an answer to the following questions:
\begin{itemize}
\item if $F$ is an immersion, what can we say about its image $F(M)$ as a subset of $N$?
\item if $F$ is a submersion, what can we say about its levelsets $f^{-1}(q) \subset M$?
\item if $F$ is a submersion, what can we say about its levelsets $F^{-1}(q) \subset M$?
\end{itemize}
And what can we say about the corresponding tangent spaces?

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