the preimage of a submanifold is locally a submanifold: Tangentspace? Let $f \colon M \to N$ be a smooth map between smooth manifold. Let $C \subset f(M)$ be a submanifold of $N$. 
Let $x \in f^{-1}(C)$ and assume that there exists a neighborhood $U \subset M$ of $x$, such that $U \cap F^{-1}(C)$ is a submanifold of $M$.
Can I then prove, that $T_x(f^{-1}(C) \cap U) = d_xf^{-1}(T_{f(x)}C)$?
It is clear, that we have $T_x(f^{-1}(C) \cap U) \subset d_xf^{-1}(T_{f(x)}C)$. So we only need to show the other inclusion. (if it is true)
 A: Let $ \ dim(M) = m \in \mathbb{N}^*$. We can view tangent vectors on $ \ x \in M \ $ as equivalence classes of curves in this manner: Let $R_x$ be the set of all (continuous) paths $ \ \gamma: J \to M \ $ such that $ \ J \subset \mathbb{R} \ $ is an open interval, $0 \in J$, $\gamma(0) = x \ $ and there exists some chart $ \ z : V \to \mathbb{R}^m \ $ around $ \ x \in V \subset M \ $ such that $ \ z \circ \gamma : J \to \mathbb{R}^m \ $ is differentiable at $0$, where $J$ is chosen small enough for we to have that $ \ im(\gamma) \subset V$, by continuity. In this way, it is simple to show that $ \ y \circ \gamma : J \to \mathbb{R}^m \ $ is differentiable at $0$, for all charts $ \ y : W \to \mathbb{R}^m \ $ around $ \ x \in W \subset M$, $\forall \gamma \in R_x$, restricting $\gamma$ to a smaller open interval, if necessary. We define an equivalence relation on $R_x$ setting $ \ \alpha \sim \beta \ $ if and only if there exists some chart $ \ z : V \to \mathbb{R}^m \ $ around $ \ x \in V \subset M \ $ such that the derivatives $ \ (z \circ \alpha)'(0) = (z \circ \beta)'(0)$, $\forall \alpha , \beta \in R_x$. Again, it is simple to show that, $\forall \alpha , \beta \in R_x$, if $ \ \alpha \sim \beta$, then $ \ (y \circ \alpha)'(0) = (y \circ \beta)'(0)$, for all charts $ \ y : W \to \mathbb{R}^m \ $ around $ \ x \in W \subset M$. Then, the relation $ \, \sim \, $ is an equivalence relation and we set $ \ T_x M = R_x/ \! \! \sim \, $. Also, for each chart $ \ z : V \to \mathbb{R}^m \ $ around $ \ x \in V \subset M \ $, we have that the map $ \ d_x z : T_x M \to \mathbb{R}^m \ $ such that, $\forall \gamma \in R_x$, $d_x z ([ \gamma ]) = [(z \circ \gamma)'(0)](1)$, is a bijection and we define a $\mathbb{R}$-vector space structure (addition and scalar multiplication) on $T_xM$ imposing that it is an isomorphism. Let $X$ be a submanifold of $M$. Because every path $ \ \gamma: J \to X \ $ is a path on $M$ too, $ \ \gamma: J \to M \ $, then it is straightforward that $ \ T_xX \subset T_xM$, $\forall x \in X$.
Let $M$ and $N$ be differentiable manifolds and $ \ f: M \to N \ $ be a differentiable function. We define the differential pushforward of $f$ at $ \ x \in M \ $ as a map $ \ d_x f : T_xM \to T_{f(x)}N \ $ such that, $\forall \gamma \in R_x$, $d_xf([\gamma])=[f \circ \gamma]$. Then, $d_xf$ is a linear map.
That being said, let $ \ [\gamma] \in T_x \big( f^{-1} [C] \cap U \big)$. Then $\gamma$ is a path $ \ \gamma : J \to f^{-1} [C] \cap U$, where $J$ is an open interval. Then $ \ d_xf([\gamma]) = [f \circ \gamma] \in T_{f(x)}N$, where $f \circ \gamma$ is a path $ \ f \circ \gamma : J \to N$. For all $ \ t \in J \ $ we have $ \ \gamma(t) \in f^{-1} [C] \cap U \subset f^{-1} [C]$. Hence $ \ (f \circ \gamma) (t) = f \big( \gamma(t) \big) \in C$. Therefore $ \ im(f \circ \gamma) \subset C \ $ and we have that $f \circ \gamma$ is a path $ \ f \circ \gamma : J \to C$. For this reason, $d_xf([\gamma]) = [f \circ \gamma] \in T_{f(x)}C \ $ and we conclude that $ \ [\gamma] \in (d_xf)^{-1}[T_{f(x)}C]$. So, we proved that $ \ T_x \big( f^{-1} [C] \cap U \big) \subset (d_xf)^{-1}[T_{f(x)}C]$.
For the other hand, let $ \ [\gamma] \in (d_xf)^{-1}[T_{f(x)}C]$. Then $ \ [\gamma] \in T_xM$, $\gamma$ is a path $ \ \gamma : J \to M \ $ and $ \ [f \circ \gamma] = d_xf([\gamma]) \in T_{f(x)}C \subset T_{f(x)}N$. Hence, $f \circ \gamma$ is a path $ \ f \circ \gamma : J \to C \ $ and we have that, $\forall t \in J$, $f \big( \gamma(t) \big) = (f \circ \gamma)(t) \in C$, which implies $ \ \gamma(t) \in f^{-1} [C]$. By continuity, we can choose $J$ small enough to have $ \ im(\gamma) \subset U$. For this reason, $\gamma(t) \in f^{-1}[C] \cap U$, $\forall t \in J$. So, we have $ \ im(\gamma) \subset f^{-1}[C] \cap U \ $ and $\gamma$ is really a path $ \ \gamma : J \to f^{-1}[C] \cap U$. Then $ \ [\gamma] \in T_x \big( f^{-1} [C] \cap U \big)$. We proved that $ \ (d_xf)^{-1}[T_{f(x)}C] \subset T_x \big( f^{-1} [C] \cap U \big)$.
Finally, we conclude that $ \ T_x \big( f^{-1} [C] \cap U \big) = (d_xf)^{-1}[T_{f(x)}C]$.
