693: For Open Subset of d1-Dimensional Euclidean C∞ Manifold, C∞ Map into d2-Dimensional Euclidean C∞ Manifold Divided by Never-Zero C∞ Map into 1-Dimensional Euclidean C∞ Manifold Is C∞

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that for open subset of $d_1$-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold, $C^{\mathrm{\infty }}$ map into $d_2$-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold divided by never-zero $C^{\mathrm{\infty }}$ map into 1-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold is $C^{\mathrm{\infty }}$

---

Topics

About: $C^{\mathrm{\infty }}$ manifold

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof

---

Starting Context

• The reader knows a definition of map from open subset of Euclidean $C^{\mathrm{\infty }}$ manifold into subset of Euclidean $C^{\mathrm{\infty }}$ manifold $C^k$ at point, where $k$ excludes $0$ and includes $\mathrm{\infty }$.

---

Target Context

• The reader will have a description and a proof of the proposition that for any open subset of the $d_1$-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold, any $C^{\mathrm{\infty }}$ map into the $d_2$-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold divided by any never-zero $C^{\mathrm{\infty }}$ map into the 1-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold is $C^{\mathrm{\infty }}$.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
${\mathbb{R}}^{d_1}$: $=\text{ the Euclidean }{C}^{\mathrm{\infty }}\text{ manifold }$
$U$: $\in \{\text{ the open subsets of }{\mathbb{R}}^{d_1}\}$
${\mathbb{R}}^{d_2}$: $=\text{ the Euclidean }{C}^{\mathrm{\infty }}\text{ manifold }$
$\mathbb{R}$: $=\text{ the Euclidean }{C}^{\mathrm{\infty }}\text{ manifold }$
$f_1$: $:U\to {\mathbb{R}}^{d_2}$, $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
$f_2$: $:U\to \mathbb{R}$, $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
//

Statements:
$\mathrm{\forall }p\in U(f_2(p)\ne 0)$
$⟹$
$f_1/f_2:U\to {\mathbb{R}}^{d_2}\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
//

---

2: Natural Language Description

For the Euclidean $C^{\mathrm{\infty }}$ manifold, ${\mathbb{R}}^{d_1}$, any open subset, $U\subseteq {\mathbb{R}}^{d_1}$, the Euclidean $C^{\mathrm{\infty }}$ manifold, ${\mathbb{R}}^{d_2}$, the Euclidean $C^{\mathrm{\infty }}$ manifold, $\mathbb{R}$, any $C^{\mathrm{\infty }}$ map, $f_1:U\to {\mathbb{R}}^{d_2}$, and any $C^{\mathrm{\infty }}$ map, $f_2:U\to \mathbb{R}$, if $f_2$ is never zero on $U$, $f_1/f_2:U\to {\mathbb{R}}^{d_2}$ is a $C^{\mathrm{\infty }}$ map.

---

3: Proof

Whole Strategy: Step 1: prove the proposition that $f_1/f_2$ is $C^1$ and the derivative is a $C^{\mathrm{\infty }}$ map divided by a never-zero $C^{\mathrm{\infty }}$ map; Step 2: prove by induction the proposition that $f_1/f_2$ is $C^n$ and the $n$-th derivative is a $C^{\mathrm{\infty }}$ map divided by a never-zero $C^{\mathrm{\infty }}$ map; Step 3: conclude that proposition of this article.

As is described in Note for the definition of map from open subset of Euclidean $C^{\mathrm{\infty }}$ manifold into subset of Euclidean $C^{\mathrm{\infty }}$ manifold $C^k$ at point, where $k$ excludes $0$ and includes $\mathrm{\infty }$, $U_p$ cited in the definition can be taken to be $U$. So, we do so hereafter.

Note that the superscripts like $f^k$ hereafter except $(f_2{)}^2$ denote the $k$-th components, not powers.

Step 1:

Let us prove the proposition that $f_1/f_2$ is $C^1$ and the derivative is a $C^{\mathrm{\infty }}$ map divided by a never-zero $C^{\mathrm{\infty }}$ map.

On $U$, ${\partial }_j(f_1/f_2{)}^k={\partial }_j({f_1}^k/f_2)=(({\partial }_j{f_1}^k)f_2-{f_1}^k{\partial }_j{f}_2)/(f_2{)}^2$, which is continuous because ${\partial }_j{f_1}^k$, $f_2$, ${f_1}^k$, ${\partial }_j{f}_2$, and $(f_2{)}^2$ are continuous and $(f_2{)}^2$ is never zero. $({\partial }_j{f_1}^k)f_2-{f_1}^k{\partial }_j{f}_2$ is $C^{\mathrm{\infty }}$, because ${\partial }_j{f_1}^k$, $f_2$, ${f_1}^k$, and ${\partial }_j{f}_2$ are $C^{\mathrm{\infty }}$. $(f_2{)}^2$ is never-zero $C^{\mathrm{\infty }}$.

So, the proposition has been proved.

Step 2:

Let us prove by induction the proposition that $f_1/f_2$ is $C^n$ and the $n$-th derivative is a $C^{\mathrm{\infty }}$ map divided by a never-zero $C^{\mathrm{\infty }}$ map.

For $n=1$, the proposition of Step 1 has proved it.

Let us suppose the proposition through $n=n^{\prime }-1$ and think of $n=n^{\prime }$.

Let the $n^{\prime }-1$-th derivative be $g_1/g_2$. $g_1/g_2$ satisfies the condition of the proposition of Step 1, so, it is $C^1$ and the derivative is a $C^{\mathrm{\infty }}$ map divided by a never-zero $C^{\mathrm{\infty }}$ map. That means that $f_1/f_2$ is $C^{n^{\prime }}$ and the $n^{\prime }$-th derivative is a $C^{\mathrm{\infty }}$ map divided by a never-zero $C^{\mathrm{\infty }}$ map.

So, the proposition has been proved.

Step 3:

The proposition of Step 2 immediately implies the proposition of this article.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>