# What are $\partial/\partial f^j$ in Jost's definition of the differential mapping?

Let $M$ be a $d$-manifold and $x_0=(x^1,x^2,\cdots, x^d)\in M$, Jost defines the tangent space at $x_0$ to be $$\{x_0\}\times \operatorname{span}\left\{\frac{\partial}{\partial x^j}\right\}.$$ So I understand these $\frac{\partial}{\partial x^j}$ to be symbols that specify the directions at $x_0$, and this seems to be a reasonable definition.

But then he defines the differential of $f:M\to N$, where $f=(f^i)$ is a differentiable function from $M$ to another manifold $N$ to be $$v_j \frac{\partial}{\partial x^j}\mapsto v_j\frac{\partial f^i}{\partial x^j}\frac{\partial}{\partial f^i},$$and this is what got me confused.

What are these $\frac{\partial}{\partial f^i}$? If we just take $N=\mathbb{R}^c$, isn't the differential be the mapping $$v_j\frac{\partial f^i}{\partial x^j}e_i$$where $e_i$ is the natural basis for $\mathbb{R}^c$?

So in my opinion these $e_i$'s should be independent of $f$ but in Jost's definition they seem to be related to the function $f$.

Where did I get it wrong? How do we actually understand these $\frac{\partial }{\partial f^i}$?

Thanks!

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The differential of $f$ at $p$ is a linear map from $T_pM$ to $T_{f(p)}N$. So it's written in terms of derivations on $N$, not on $M$. – user31373 Sep 21 '12 at 2:16
@LVK What do yo mean? I do not quite understand. – Hui Yu Sep 21 '12 at 2:37
The components of $f$ are with respect to some coordinate system $y^i$, and the author is abusing notation by writing $\frac{\partial}{\partial f^i}$ for $\frac{\partial}{\partial y^i}$. But it's perfectly understandable: this way, it looks like the pushforward is nothing more than the chain rule. – Zhen Lin Sep 21 '12 at 2:40

Suppose we have coordinates $(x^i)$ in a neighborhood of point $p$ and also coordinates $(y^{\lambda})$ in a neigborhood of point $f(p)$, so locally map $f$ can be identified with $y^{\lambda} = f^{\lambda}(x^i)$. As @LVK pointed the differential $Tf$ at $p$ is the linear map $T_{p}f \colon T_p M \to T_{f(p)} N$ that maps every coordinate vector field $\partial_i = \tfrac{\partial}{\partial x^i}|_{p}$ to $\frac{\partial f^{\lambda}}{\partial x^i} \frac{\partial}{\partial y^{\lambda}}|_{f(p)}$ and extended by linearity. These $\tfrac{\partial}{\partial f^{\lambda}}$ are just a shorthand notation for the fields $\frac{\partial}{\partial y^{\lambda}}|_{f(p)}$.
In fancier terms the differential of $f$ can be seen as a section of $T^* M \otimes f^* TN$ given by $$df = \frac{\partial f^\lambda }{\partial x^i} dx^i\otimes \frac{\partial}{\partial y^\lambda}$$ where the Einstein summation is assumed.
The fields $\tfrac{\partial}{\partial f^{\lambda}} \Big|_{p} = \left( \frac{\partial}{\partial y^{\lambda}} \right)_{f} \Big|_{p}= \frac{\partial}{\partial y^{\lambda}} \Big|_{f(p)}$ are a frame in the pullback bundle $f^* TN$