# Why is this differential injective? Lee Smooth Manifolds Proposition 5.3

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?

• Try and show that if any linear map has a left inverse, then it is injective. Feb 27 at 18:20

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 $$g = \operatorname{d}(\pi_M)_{(x,f(x))}$$.
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.