Inverse image of a sub manifold - Transversal intersection

Solution 1:

“$ f $ is transverse to $ S $” means that $ {T_{f(x)}}(M) = {(\mathrm{d} f)_{x}}({T_{x}}(N)) + {T_{x}}(S) $ for any $ x \in N $.

Consider a chart $ \varphi: U \to \Omega $ of $ M $ that is adapted to $ S $ and centered at $ f(x) $. “Adapted” means that $ \varphi[U \cap S] = \Omega \cap (\mathbb{R}^{k} \times \{ O \}) $, where $ k $ denotes the dimension of $ S $.

Observe that $ {T_{f(x)}}(S) = {(\mathrm{d} \varphi^{- 1})_{0}}(\operatorname{Span}(e_{1},\ldots,e_{k})) $.

Let $ \psi = (\varphi^{k + 1},\ldots,\varphi^{n}) $; we will prove that $ O $ is a regular value of $ \psi \circ f $. As $ \varphi $ is a diffeomorphism, \begin{align} \mathbb{R}^{n} & = {(\mathrm{d} \varphi)_{f(x)}}({T_{f(x)}}(M)) \\ & = {(\mathrm{d} \varphi)_{f(x)}}({(\mathrm{d} f)_{x}}({T_{x}}(N)) + {T_{x}}(S) \\ & = {(\mathrm{d} \varphi)_{f(x)}}({(\mathrm{d} f)_{x}}({T_{x}}(N)) + {(\mathrm{d} \varphi)_{f(x)}}({T_{f(x)}}(S)) \\ & = {(\mathrm{d} \varphi)_{f(x)}}({(\mathrm{d} f)_{x}}({T_{x}}(N)) + \mathbb{R}^{k} \times \{ O \}, \end{align} meaning that $ \{ O \} \times \mathbb{R}^{n - k} \subseteq {(\mathrm{d} \varphi)_{f(x)}}({(\mathrm{d} f)_{x}}({T_{x}}(N)) $. In particular, $$ \mathbb{R}^{n - k} \subseteq {(\mathrm{d} \psi)_{f(x)}}({(\mathrm{d} f)_{x}}({T_{x}}(N)), $$ so $ O $ is a regular value.

Consequently, we deduce that $ {(\psi \circ f)^{- 1}}[O] = {f^{- 1}}[S \cap U] $ is a sub-manifold of $ N $ of co-dimension $ n - k $. To conclude, simply cover $ S $ with adapted charts $ U_{i} $ and then take the inverse image of the intersection. This will prove that $ {f^{- 1}}[S] $ can be covered by charts $ {f^{- 1}}[S \cap U_{i}] $, so it will be a sub-manifold.