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For a Riemannian manifold $M$, and each $x\in M$, we can define the exponential map

$$\exp_x:T_xM\to M.$$

Then for each vector $v\in T_x M$, we have the differential

$$(d\exp_x)_v:T_vT_xM \to T_{\exp_x(v)}M.$$

Why do people say $(d\exp_x)_v$ is the identity map? I understand that we can identify $T_vT_xM$ with $T_xM$ by a translation $u\mapsto u-v$. But $T_xM$ and $T_{\exp_x(v)}M$ are completely different vector spaces!

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    Are you sure people say it in that generality? With the natural identification $T_0T_xM \cong T_xM$, we get $(d\exp_x)_0 = \operatorname{id}_{T_xM}$, but as written it doesn't look very natural. – 2017-01-27
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    As far as I know "people say" this only in the context of the proof of the theorem that $\exp_p$ is a local diffeomorphism when restricted to a small ball centered at the origin in $T_pM$. If you know any other context, please, provide a() reference(s). – 2017-01-27
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    Thanks Daniel Fischer and Moishe Cohen, I'm convineced that I had misunderstood something, you guys are correct. – 2017-01-29

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