If $U\subseteq \mathbb{R}^n$ and $V\subseteq \mathbb{R}^m$, let's define immersions $f\colon U\longrightarrow \mathbb{R}^n$ and $g\colon V\longrightarrow \mathbb{R}^m$, if I define $$h\colon U\times V\longrightarrow \mathbb{R}^{m+n}$$ With $h((u,v))=(f(u),g(v))$, then it is an immersion too? Why?
Asked
Active
Viewed 297 times
1 Answers
4
It is, because $ Dh = Df \oplus Dg $ and the direct sum of two injective linear maps is injective. More explicitly,
$$ Dh = \begin{pmatrix} \frac{\partial f}{\partial u} & \frac{\partial f}{\partial v} \\ \frac{\partial g}{\partial u} & \frac{\partial g}{\partial v} \end{pmatrix} = \begin{pmatrix} Df & 0 \\ 0 & Dg \end{pmatrix}$$
where $Df$ is the partial derivatives matrix of $f$ with respect to the $u^i$-variables, $Dg$ is the partial derivatives matrix of $g$ with respect to the $v^i$ variables, $\frac{\partial f}{\partial v}$ is the partial derivatives matrix of $f$ with respect to the $v^i$ variables (which vanishes, as $f$ is independent of $v^i$), etc.
levap
- 65,634
- 5
- 79
- 122
-
Why it lets that $Dh=Df\oplus Dg$? It looks obvious, but I don't know the reason. – iam_agf Nov 23 '16 at 01:48
-
@MonsieurGalois: I've added some details. – levap Nov 23 '16 at 01:51