Let $X\subseteq\mathbb{R}^k$, $Y\subseteq\mathbb{R}^m$ and $f\colon X\rightarrow Y$. We want to declare when such a map ought to be called differentiable and there are two possible notions, both of which may feel natural. We call $f$ differentiable if, for each $x\in X$, there is an open neighborhood $U$ of $x$ and a differentiable (in the usual sense) extension $\hat{f}\colon U\rightarrow\mathbb{R}^m$ of $f$. We can also do the same thing, but additionally require that $\hat{f}(U)\subseteq Y$. Both of these definitions yield a notion of morphisms in a category where objects are subsets of …
$T_p(\mathbb{R}^n)\cong D_p(\mathbb{R}^n)$
All I want to show is that the map:
$\phi: T_p(\mathbb{R}^n)\rightarrow D_p(\mathbb{R}^n)$ , given by $\phi(v_p)=D_v=\sum_k v^k \frac{\partial}{\partial{x}^k}|_p.$ is surjective.
Let $D$ be a derivation at $p$. Let $[(f,V)]$ be a germ at p (this exi...
On a set A ,define a binary operator * such that
$∀x,y ∈ A$,
we have
$x*(x*y)=y$
$(y*x)*x=y$
Is * associative?
Note:* is commutative because
We have
$x=x$
$<=>y*(y*x)=[x*(y*x)](y*x)$
$<=>[y*(y*x)]*(y*x)=\{[x*(y*x)](y*x)\}(y*x)$
$<=>y=x*(y*x)$
$<=>x*y=x*(x*(y*x))$
$<=>x*y=y*x$
Could yo...
