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Let $M$ be a smooth manifold, and $E\to M$ a smooth vector bundle. Then if we have a smooth morphism $f:N\to M$ of smooth manifolds, then we obtain the following commutative diagram, $$\require{AMScd} \begin{CD}f^*E @>>> E \\ @VVV @VVV \\ N @>f>> M \end{CD}$$ where $f^*E$ is the pullback of $E$. Therefore, if we have some bundle metric $g\in E^*\otimes E^*$, we can define the pullback bundle metric $\tilde g:=f^*g\in f^*E^*\otimes f^*E^*$ in the natural way, and for any connection $$\nabla:\Gamma(\mathrm TM)\otimes\Gamma(E)\to\Gamma(E)$$ we can define the pullback connection $$f^*\nabla:\Gamma(\mathrm TN)\otimes\Gamma(f^*E)\to\Gamma(f^*E)$$ as follows:

Let $(x^i)$ be coordinates on $N$ mapping into coordinates $(y^j)$ on $M$, and let $E$ be locally trivial with basis $(e_k)$ on $M$, then if $\nabla_{\partial/\partial y^i}e_j=\Gamma_{ij}^k e_k$, then if we set $$\tilde\Gamma_{ij}^k=\frac{\partial f^l}{\partial x^i}\Gamma_{lj}^k\circ f$$ then we can define $\tilde\nabla$ by letting $\tilde\nabla_{\partial/\partial x^i} e_j=\tilde\Gamma_{ij}^k$.

Now, suppose $\nabla$ is compatible with $g$, that is to say, for $a,b\in\Gamma(E)$ and $X\in\Gamma(\mathrm T M)$, $$X\langle a,b\rangle = \langle\nabla_Xa,b\rangle+\langle a,\nabla_Xb\rangle$$ then I wish to prove that for $a,b\in\Gamma(f^*E)$ and $X\in\Gamma(\mathrm T N)$, $$X\langle a,b\rangle = \langle\tilde\nabla_X a,b\rangle+\langle a,\tilde\nabla_Xb\rangle$$ but is this true? I've been having trouble proving this even for $X=\frac{\partial}{\partial x^i}$ and $a=e_j$, $b=e_k$. In this case, I obtain that

$$\frac{\partial}{\partial x^i}\langle e_j,e_k\rangle = \frac{\partial g_{jk}}{\partial x^i} = \frac{\partial g_{jk}}{\partial y^l}\frac{\partial f^l}{\partial x^i}$$ and

$$\langle\tilde\nabla_{\partial/\partial x^i}e_j,e_k\rangle + \langle e_j,\tilde\nabla_{\partial/\partial x^i}e_k\rangle =\tilde\Gamma_{ij}^l g_{lk} + \tilde\Gamma_{ik}^l g_{lj} = \Gamma_{mj}^l\frac{\partial f^m}{\partial x^i}g_{lk} + \Gamma_{mk}^l\frac{\partial f^m}{\partial x^i}g_{lj} $$ but I don't see why these two expressions should be equal.

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    $\begingroup$ Note that it is wrong to think of $\nabla$ as a map $\Gamma(TM)\otimes\Gamma(E)\to\Gamma(E)$, as it is only tensorial in the $\Gamma(TM)$ part. $\endgroup$ – Amitai Yuval Jun 18 '17 at 2:52
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The answer is yes - the pullback connection is indeed compatible with the pullback metric. As always, there are many different ways to show, or even understand this claim. The following is one of those ways.

Every connection is determined by its parallel transport, and a connection is compatible with a given metric if and only if the associated parallel transport preserves the metric. Hence, we wish to show that if parallel transport with respect to $\nabla$ preserves $g$, then parallel transport with respect to $f^*\nabla$ preserves $f^*g$. This follows immediately from the following characterization of parallel transport with respect to $f^*\nabla$:

Let $\gamma:[0,1]\to N$ be a smooth path. Then we have $f\circ\gamma:[0,1]\to M$, and the fibers of $f^*E$ over the endpoints of $\gamma$ are identified with the fibers of $E$ over the endpoints of $f\circ\gamma$. Parallel transport with respect to $f^*\nabla$ along $\gamma$ is identical to parallel transport with respect to $\nabla$ along $f\circ\gamma$. (This is actually one of the ways to define the pullback connection).

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You have already obtained the answer. From your last expression using the identity $$g_{kl}\Gamma^{l}_{mj}+g_{jl}\Gamma^{l}_{mk}=\partial_{y^m}g_{jk}$$ one gets $$(g_{kl}\Gamma^{l}_{mj}+g_{jl}\Gamma^{l}_{mk}){\partial f^m\over\partial x^i}={\partial f^m\over\partial x^i}{\partial g_{jk}\over\partial y^m}$$ Which is the expected answer. You can check the identity by using the standard form for Christoffels $$\Gamma^{i}_{jk}={g^{il}\over 2}({\partial g_{lk}\over\partial y^j}+{\partial g_{lj}\over\partial y^k}-{\partial g_{jk}\over\partial y^l})$$

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