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It has come up in my differential geometry class that if $f:\mathbb{R}\to\mathbb{R}$ is a smooth function, then we can 'identify' the pushforward $df_p:T_p \mathbb{R}\to T_f(p)\mathbb{R}$ with $f'$.

My reasoning:

Let $Id_\mathbb{R}:\mathbb{R}\to\mathbb{R}$ denote the identity on $\mathbb{R}$. Then $\{\frac{d}{dt}|_{p}$} is a basis for $T_p\mathbb{R}$. Then $df_p(\frac{d}{dt}|_{p})(Id_\mathbb{R})=\frac{d}{dt}|_{t_0} (f\circ {Id}_\mathbb{R})=\frac{d}{dt}|_{t_0}=f'(p)$

Jungleshrimp
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  • You need to write $df_p$ and $f'(p)$. And $df_p\colon T_p\Bbb R\to T_{f(p)}\Bbb R$, right? Now be careful with your last line. What function of $t$ are you differentiating and why is $t=t_0$? – Ted Shifrin Apr 04 '22 at 00:16
  • This is basically true. The main thing to keep in mind is just the notation. In general, you can think of the derivative of a map $f\colon \Bbb R^n\to \Bbb R^m$ as a map from $\Bbb R^n$ to the space of $m\times n$ matrices. When $m=n=1$, these matrices just have one entry, so you can indeed think of $f'$ as another map $\Bbb R\to\Bbb R$. – pancini Apr 04 '22 at 00:22
  • Is this the same? https://math.stackexchange.com/questions/193560/difference-between-pushforward-and-differential – Andrew D. Hwang Apr 05 '22 at 01:39

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