I want to ask a question about the pullback of the Riemann metric with the map being the exponential map.
Given a Riemannian manifold $(M,g)$, and an exponential map $f_\alpha: B_1 \rightarrow M$ which is given by $$f_\alpha(x):=\exp _{x_0}(\alpha x),$$ where $\exp _p$ is the exponential map at $p$, then define $$h_\alpha:=\alpha^{-2} f_\alpha^* g,$$ which is a metric on $B_1$.
How to prove that $$ h_{\alpha, i j}=\alpha^{-2} g_{i j}, $$ when $M$ is $S^2$, we can do some calculations:
Let $x_0$ be the north pole of $S^2$. Write $v \in T_{x_0} S^2$ as $v=t(\cos \theta, \sin \theta)$, the $(\phi, \theta)$ coordinates of $\exp _{x_0}(\alpha v)$ are $\phi=\alpha t$, $\theta=\theta$.
Take the induced metric on $S^2$ from pullback of metric on $\mathbb{R}^3$, so the metric is $$d s^2=d \phi \otimes d \phi+\sin ^2 \phi d \theta \otimes d \theta,$$ so $$f_{\alpha}:(t,\theta) \rightarrow (\phi,\theta),$$ with $(t,\theta)$ being $(x_1,x_2)$ and $(\phi,\theta)$ being $(y_1,y_2)$, and $$ \left(f_{\alpha}^* g\right)_{\mu \nu}=\frac{\partial y^\alpha}{\partial x^\mu} \frac{\partial y^\beta}{\partial x^\nu} g_{\alpha \beta} $$ then $$\left(f_\alpha^* g\right)_{1 1}=\alpha^2 g_{11}, \left(f_\alpha^* g\right)_{1 2}=\left(f_\alpha^* g\right)_{2 1}=0, \left(f_\alpha^* g\right)_{22}=g_{22},$$ so $$h_\alpha=\alpha^{-2}\left(\alpha^2 d t \otimes d t+\sin ^2(\alpha t) d \theta \otimes d \theta\right) ,$$ so $h_{\alpha, i j}=\alpha^{-2} g_{i j}$ as desired.
But what about when $M$ is not something we could calculate? I have no idea about this. This seems to be fundamental but I just don't know.
Update: Is it because that given a map $M \rightarrow \tilde{M}$, the pushforward of basis vectors of $T_p(M)$ are indeed the basis vectors of $T_q(\tilde{M})$?