# $df_x$ maps $N(S.X)$ isomorphically

This is actually just a linear algebra problem, but this text is taken from Guillemin and Pollack, Differential Topology on page 100.

Real Problem: Let $f: X \to Y$ be smooth, $S = f^{-1}(Z)$ where $T_x(S) = (df)^{-1}(T_z(Z))$, $f(x) = z$ and $N(S;X)$ is the orthogonal complement to $T_x(S)$. Then since $T_x(S)$ contains the whole kernel of $df_x$, $df_x$ maps $N(S;X)$ isomorphically into $df N(S;X)$.

If anyone is confused as to why $T_x(S)$ contains the full kernel, it is because $f^{-1}(z) \subset f^{-1}(Z)$ and it is shown previously that $\ker df = T_x(f^{-1}(z))$. Also just a side note, does the tangent space come equipped with an inner product? How do they talk about orthogonal complements without it?

What this problem actually say in Linear Algebra: We linear map $g: V \to W$ where $V = S \oplus S^{\perp}$. $\ker g \subset S$, then $g : S^{\perp} \to g(S^\perp)$ isomorphically.

I can't quite pinpoint why $g|_{S^{\perp}}$ is isomorphism. Is it because $\ker g|_{S^{\perp}} = 0$ since the whole kernel is contained in $S$, but $S \cap S^{\perp} = \{0\}?$

Sorry this post appeared longer than it should have.

• A typo in the translation to linear algebra, $\text{ker } g \subset V$ should be $\text{ker } g \subset S$. – hunter Aug 10 '18 at 3:18
• As to the inner product, if I recall the setup in Guillemin and Pollack's textbook correctly, each tangent space lives in some giant $\mathbb{R}^N$ and then you can just restrict the standard inner product to it. – hunter Aug 10 '18 at 3:25
• @hunter, I think they defined it as the image of the chart $\phi$. – Hawk Aug 10 '18 at 3:34
• Fixed thank you. – Hawk Aug 10 '18 at 3:38

To say that $g|_{S^\perp}: S^\perp \to g(S^\perp)$ is an isomorphism is just to say that it's injective, since it's onto its image by definition.
But if $v \in \ker(g|_{S^\perp})$, then $v \in S \cap S^\perp = \{0\}$.