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Let $M,N$ be two differentiable manifolds and $f: M\to N$ be a smooth map. Then the tangent map $Tf: TM\to TN$ is a homomorphism of vector bundles.

  1. If we assume that $f$ is injective, whether can we conclude or not that $Tf$ induces a pull-back, that is, $TM=(Tf)^*TN$?

  2. If we assume that for any $p\in M$, $Tf_p$ is injective, whether can we conclude or not that $Tf$ induces a pull-back, that is, $TM=(Tf)^*TN$?

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    $\begingroup$ do you assume that $\mathrm{dim} M = \mathrm{dim} N$? otherwise $TM$ and $F^*TN$ have different fibers $\endgroup$ – Andrey Ryabichev Aug 29 '16 at 12:01
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Well, in general $Tf$ induces a push forward between $TM$ and $TN$. Now, if you take $T^*N$ and $T^*M$, then $(Tf)^* : T^*N \to T^*M$ is the natural pull back. If $\dim TM < \dim TN$ then I don't know how you can induce a pull back as $TM = (Tf)^* TN$ for vectors that lie outside the image of $Tf$. However, you can induce a new bundle $f^*(TN)$ on $M$ by the construction $$f^*(TN) =\Big\{(x, v) \in M \times TN \,\, : \,\, f(x) = \pi_N(v) \, \Big\},$$ which is a bundle over $M$ with fibers $T_{f(x)}N$. Here $\pi_N : TN \to N$ is the bundle projection and the bundle map $\pi_{f^*(TN)} : f^*(TN) \to M$ is defined as $\pi_{f^*(TN)}(x,v) = x$ for $(x,v) \in f^*(TN)$. This however does not require for $f$ to be injective. Also $f$ injective does not necessarily imply $Tf$ injectivity.

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