This question actually comes from the question I asked before:
Derivative map of the diagonal inclusion map on manifolds
And I repeat it as follows:
Let $f: X\longrightarrow X\times X$ be the mapping $f(x)=(x,x)$. Check that $df_x(v)=(v,v)$. Here $X\subset \mathbf R^m$ is a manifold.
I thought of a new way of doing it without going through the painful commutative squares. But it was so simple that I suspect it might be wrong.
I was thinking of this: Pick a vector $v$ in the tangent plane of $X$ at point $x$($v\in T_x(X)$). Then according to the definition of derivative map, we can compute
$$\lim_{t\to 0}\frac{f(x+tv)}{t}=\lim_{t \to 0} \frac{(x+tv,x+tv)-(x,x)}{t}=(v,v)$$
Therefore the derivative map $df_x(v)=(v,v)$.
I was wondering if there a loophole in my argument? It just looks so much simpler...
Also just for everyone's convenience, so that you don't have to cross reference, I post the accepted answer for my previous question as follows(it's given by Giuseppe):
Hoping it helps you, I am expanding A.Bellmunt's comment.
Let be $x$ a point of $X$, a submanifold of $\mathbb R^n$, and $\phi:X\to\mathbb R^m$ a local coordinate chart centered at $x$ (i.e. $\phi(x)=0$).
Therefore $\phi\times\phi:X\times X\to\mathbb R^m\times\mathbb R^m$ is a local coordinates chart centered at $(x,x)$ (i.e. $(\phi\times\phi)(x,x)=(0,0)$).
Now we get the local expression $f=(\phi\times\phi)^{-1}\circ \widetilde{f}\circ\phi$, where $\widetilde{f}$ is the linear map $$\widetilde{f}:u\in\mathbb R^m\to(u,u)\in\mathbb R^m\times\mathbb R^m.$$ Therefore:
- $\widetilde f$ is linear, so it coincides with $d_0\widetilde f$, and
- if $v\in T_xM \overset{d_x\phi}{\longrightarrow}\tilde v\in\mathbb R^m$, then $(v,v)\in T_{(x,x)}X\times X\overset{d_{(0,0)}(\phi\times\phi)}{\longrightarrow}(\tilde v,\tilde v)\in\mathbb R^m\times\mathbb R^m$,
and by 1. and 2. we get immediately the searched expression of $d_xf$, i.e.: $$\begin{array}{ccc} v\in T_xM&\overset{d_xf}{\longrightarrow}&(v,v)\in T_{(x,x)}X\times X\\ \downarrow d_x\phi&&\downarrow d_{(x,x)}(\phi\times\phi)\\ \tilde v\in\mathbb R^m&\overset{d_0\widetilde f}{\longrightarrow}&(\tilde v,\tilde v)\in\mathbb R^m\times\mathbb R^m\end{array}$$
Unfortunately, your attempt only makes sense once you're working in local coordinates. When you pick $v\in T_xX$, the point $x+tv$ rarely lives in $X$, so you cannot make sense of $f(x+tv)$.