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I see that in other posts, people state $$dA = -I_{n\times n}$$ and I was wondering how would I prove this explicitly using definition of differential.

First question, is there a general formula for $df$ when $f:M\rightarrow N$ is a smooth map between smooth manifolds?

I know when $f:M \rightarrow \mathbb{R}$, fix a chart around $p\in M$, we have the formula $$df|_p = \frac{\partial f}{\partial x_i} dx_i|_p$$ which means $df$ is a cotangent vector (or called one form). Now if $f: M\rightarrow \mathbb{R}^k$, by calculation, I think $df$ would be a matrix $$df_{ij} = \frac{\partial \pi_i\circ f}{\partial x_j},$$ is there a special name for this $df$? And should I think of it as a tensor product between a cotangent vector on $M$ and a cotangent vector on $R^k$ $$df_{ij} = \frac{\partial \pi_i\circ f}{\partial x_j}dr_i\otimes dx_j?$$ (so that I can generalize this to $f:M\rightarrow N$, but this would be a (2,0) tensor on two different manifolds...)

Here we have the antipodal $A:S^n \rightarrow S^n$, suppose we use the standard stereographic projection as our coordinate system, then the coordinate map $\phi$ on $S^n/\{n\}$ would be $$(x_1, \cdots, x_{n+1}) \mapsto \left(\frac{x_1}{1-x^n},\cdots \frac{x_n}{1-x^{n+1}}\right)$$ and its inverse $$(y_1, \cdots ,y_{n}) \mapsto \left(\frac{2y_1}{1+|y|^2},\cdots \frac{-1+|y|^2}{1+|y|^2}\right)$$ is there a formula for $dA$? Does it have something to do with $$"\frac{\partial A}{\partial x_i} "= \frac{\partial \phi\circ A\circ \phi^{-1}}{\partial y_i} = \frac{\partial}{\partial y_i}\left[ \frac{y}{|y|^2}\right]?$$

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    It's pretty obvious that the antipodal map ($x \mapsto -x$) on $\Bbb R^{2n+1}$ has derivative $-Id$ (its a linear map so it should be its own derivative). So in some sense the thing you wrote is true. I don't think it makes any intrinsic sense to say the differential of the antipodal map is -Id as the induced map on tangent spaces doesn't fix any points. In general between two different choices of charts the only thing that will remain consistent is the differentials rank. – 2017-02-14
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    @PVAL: Since $T_pS^n=T_{-p}S^n$ ($=$, not $\cong$), it actually does make sense and is right! :) – 2017-02-15
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    @TedShifrin Note the word intrinsic in my comment. My point is that while this is true if you consider $T_p S^n$ as subspaces of $\Bbb R^{2n+1}$, but that depends on the choice of embedding. Under the standard embedding $T_pS^n$ and $T_{-p}S^n$ get identified to the same subspace but they are NOT the same vector space. – 2017-02-16

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