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I'm trying to understand the following line in the proof to Proposition 5.3 in john M. Lee's 'Introduction to Smooth Manifolds':

"Because the projection $\pi_M : M \times N \rightarrow M$ satisfies $\pi_M \circ \gamma_f(x) = x$ for each $x \in U$ (where $U$) is an open subset of $M$, so the composition $d(\pi_M)_{(x, f(x))} \circ d\gamma_x$ is the identity map on $T_x M$ for each $ \in U$. Thus $d(\gamma)_x$ is injective."

  This is a silly question (apologies in advance), but why does the composition being the identity imply that $d(\gamma)_x$ is injective?

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    $\begingroup$ Try and show that if any linear map has a left inverse, then it is injective. $\endgroup$
    – Chris
    Feb 27 at 18:20

2 Answers 2

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This is true in general: If $f\colon X \to Y$ and $g\colon Y \to X$ are two functions satisfying

$$ g\circ f = \operatorname{id}_X, $$ then $f$ is injective. If $x,y \in X$ are such that $f(x) = f(y)$, then of course it is also true that $g(f(x)) = g(f(y))$ and by the above identity, $x = y$ which shows injectivity. In your case, $f = \operatorname{d}\gamma_x$ and $g = \operatorname{d}(\pi_M)_{(x,f(x))}$.

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Noam Szyfer's answer is fine, but here is another approach based on the fact that $\mathrm{d}\pi_M$ and $\mathrm{d}\gamma_x$ are linear maps. It is injective if and only if its kernel is trivial, i.e. $\ker\mathrm{d}\gamma_x = \{0\}$. It can be proved by contradiction.

Let $x \in M$ and let's assume that $\exists v \in T_xM$ be such that $v \neq 0$ but $\mathrm{d}\gamma_x(v) = 0$. Then, $(\mathrm{d}\pi_M \circ \mathrm{d}\gamma_x)(v) = \mathrm{d}\pi_M(0) = 0$, hence $v \in \ker (\mathrm{d}\pi_M \circ \mathrm{d}\gamma_x) = \ker\mathrm{id}_{T_xM} = \{0\}$ and finally $v = 0$, which permits to conclude ab absurdo.

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