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I'm trying to prove $$ \text{the image of }d_{p}f = \big\{\alpha'(t_{0}) \; \big\vert\; \text{for all smooth }\alpha:((t_{0}-\epsilon), (t_{0}+\epsilon)) \rightarrow M \text{ such that } \alpha(t_{0})= f(p) \big\} $$

For $\quad f:U\rightarrow M$

So far I have one direction:

If $v \in V_{f(p)}$ then $v=\alpha '(t_{0})$ for some smooth $\alpha:((t_{0}-\epsilon , t_{0}+\epsilon)) \rightarrow M$ such that $\alpha(t_{0})=f(p)$. $\alpha'(t_{0})=d_{t_{0}}\alpha(1), d_{t_{0}}\alpha=d_{(f^-1 \cdot \alpha)}f_{(t_{0})} \circ d_{t_{0}}(f^-1 \cdot \alpha)=d_{p}f(d_{t_{0}}(f^-1 \cdot \alpha)(1))=d_{p}f \circ d_{t_{0}}(f^-1 \cdot \alpha)$ So, $\alpha '(t_{0}) \in \text{ the image of } d_{p}f$.

But I'm having trouble showing the other direction.

bphi
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1 Answers1

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One can define a tangent space at a point as the vector space of all derivatives of curves through that point. Thus if $U$ is a (subset of) an n-dimensional manifold, then $T_pU=\{ v \in \mathbb{R}^n~|~v=c'(t_0)\}$ where $c:\mathbb{R}\rightarrow U$ is a curve on $U$ with $c(t_0)=p$.

By definition, for some map $f:U\rightarrow M$, the image of $df$ at some point $p$ is then given by $\{w\in \mathbb{R}^m~|~w=(f\circ c)'(t_0)\}$ where this time $f(c(t_0))=f(p)$.

Thus if you define your $\alpha$'s as $\alpha:= f\circ c:\mathbb{R}\rightarrow M$ restricted to the interval $(t_0-\epsilon,t_0+\epsilon)$, then you still get the whole image of $df_p$ because the derivative only depends on the curve in the vicinity of the point (which completes one direction of the proof) and the set $\alpha$'s are indeed all smooth curves you can define because you started out with all possible curves $c$ and any change along $f$ can be reparameterised by another $c$ (other direction of the proof). Thus you should obtain what you claimed at the beginning.

exchange
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