Relationship between the covector $(df)_p$ and the differential $df_p$ of $f$ at $p$

Question

Let $M$ be a smooth manifold and let $f\in C^\infty(M)$. The differential of $f$ at $p\in M$ is the linear map $df_p:T_p M\to T_{f(p)}\mathbb{R}$ defined by $$df_p(v)(g)=v(g\circ f),$$ where $v$ is a derivation in $T_p M$ and $g\in C^\infty(\mathbb{R})$. On the other hand, given $f\in C^\infty(M)$, we can obtain a smooth covector field $df$ on $M$ by defining $$(df)_p(v)=vf$$ for $p\in M$ and $v\in T_p M$. Some people claim that $(df)_p$ and $df_p$ are actually the same thing, by identifying $T_{f(p)}\mathbb{R}$ with $\mathbb{R}$. But I would like to cast a doubt on it. Let $g\in C^\infty(\mathbb{R})$. If the claim is true, we should be able to see that $$D_{(df)_p(v)}\Big|_{f(p)}g:=\frac{d}{dt}\Big|_{t=0}g(f(p)+t(df)_p(v))=v(g\circ f),$$ in which I try to go through $\mathbb{R}\cong\mathbb{R}_{f(p)}\cong T_{f(p)}\mathbb{R}$. But I ended up with $$\frac{d}{dt}\Big|_{t=0}g(f(p)+t(df)_p(v))=g'(f(p))(vf).$$ What seems to be the problem in between my calculation? Thank you.

Answer 1

Although I am still unable to dispel the doubt, $df_p$ does have something to do with $(df)_p$. In fact, I would now like to show that $\forall v\in T_p M$, $$df_p(v)=(df)_p(v)\frac{d}{dt}\Big|_{f(p)},$$ where $\frac{d}{dt}\Big|_{f(p)}$ is the coordinate basis for $T_{f(p)}\mathbb{R}$. Since $df_p(v)\in T_{f(p)}\mathbb{R}$, we can write $df_p(v)=k\frac{d}{dt}\Big|_{f(p)}$ for some $k\in\mathbb{R}$. Applying both sides to the identity map $\mathrm{Id}_\mathbb{R}$, we see $$k=df_p(v)(\mathrm{Id}_\mathbb{R})=v(\mathrm{Id}_\mathbb{R}\circ f)=vf.$$ The end product is none other than the number $(df)_p(v)$!