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Piotr Hajlasz
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1. If we do not assume that the diffeomorphism is identity outside a small ball, the answer is no. If $F$ preserves orientation but $F'$ reverses orientation, then there is no isotopy between $F$ and $F'$, but even if both diffeomorphisms preserve orientation there are counterexamples.

Here is one. Take $M=\mathbb{S}^1\times\mathbb{S}^1$ and $F:M\to M$ to be the identity. Let $F':M\to M$, $F'(z_1,z_2)=(z_2,\overline{z}_1)$ be the diffeomorphism that "changes the circles" (we take the complex conjugate - reflection in the $y$-axis to make sure that $F'$ preserves orientation). Fix $p=q=(1,1)\in\mathbb{S}^1\times\mathbb{S}^1$. Then $F(p)=F'(p)=q$. But there is no isotopy that would deform $F'$ to $F$, because $F$ and $F'$ are not homotopic. Indeed, $F(\mathbb{S}^1\times \{1\})=\mathbb{S}^1\times \{1\}$ but $F(\mathbb{S}^1\times \{1\})=\{1\}\times\mathbb{S}^1$ give two different elements in $\pi_1(M)$.

2. The answer is yes if $F$ is identity outside a small ball, but the isotopy is not smooth at the endpoint. We can assume that $F:\mathbb{B}^n\to\mathbb{B}^n$ is identity near the boundary, $p=0$ and $q\in \mathbb{B}^n$. According to our assumptions $F':\mathbb{B}^n\to\mathbb{B}^n$ be a diffeomorphism which is identity near the boundary and $F'(0)=q$. Then $(F')^{-1}\circ F$ maps $0$ to $0$. And $$ [0,1]\ni s \to F'(s((F')^{-1}\circ F)(\frac{x}{s})) $$ gives an isotopy between $F$ when $s=1$ and $F'$ when $s=0$. Moreover $p=0$ is mapped to $q$ for all $s\in [0,1]$. This isotopy is a continuous path of diffeomorphisms, but it is not smooth at $s=0$.

3. There are orientation preserving diffeomorphisms of $\mathbb{S}^6$ that are not isotopic to identity map. My understanding is that here we mean a smooth isotopy so the construction from the part 2 does not apply. Indeed, exotic $7$-spheres are obtained by gluing together two balls $\mathbb{B}^7$ along a diffeomorphism $F:\mathbb{S}^6\to\mathbb{S}^6$, see https://en.wikipedia.org/wiki/Exotic_sphere. If $F$ is smoothly isotopic to identity, then we obtain a standard sphere. Note that every orientation preserving diffeomorphism of $\mathbb{S}^6$ is isotopic to a diffeomorphism that is identity on the half sphere because near any point the difeeomorphism is close to a tangent map so this example applies to your question.

Remark. Regarding the construction in part 3, it is not my area of expertise to any comments are welcome.

1. If we do not assume that the diffeomorphism is identity outside a small ball, the answer is no. If $F$ preserves orientation but $F'$ reverses orientation, then there is no isotopy between $F$ and $F'$, but even if both diffeomorphisms preserve orientation there are counterexamples.

Here is one. Take $M=\mathbb{S}^1\times\mathbb{S}^1$ and $F:M\to M$ to be the identity. Let $F':M\to M$, $F'(z_1,z_2)=(z_2,\overline{z}_1)$ be the diffeomorphism that "changes the circles" (we take the complex conjugate - reflection in the $y$-axis to make sure that $F'$ preserves orientation). Fix $p=q=(1,1)\in\mathbb{S}^1\times\mathbb{S}^1$. Then $F(p)=F'(p)=q$. But there is no isotopy that would deform $F'$ to $F$, because $F$ and $F'$ are not homotopic. Indeed, $F(\mathbb{S}^1\times \{1\})=\mathbb{S}^1\times \{1\}$ but $F(\mathbb{S}^1\times \{1\})=\{1\}\times\mathbb{S}^1$ give two different elements in $\pi_1(M)$.

2. The answer is yes if $F$ is identity outside a small ball, but the isotopy is not smooth at the endpoint. We can assume that $F:\mathbb{B}^n\to\mathbb{B}^n$ is identity near the boundary, $p=0$ and $q\in \mathbb{B}^n$. According to our assumptions $F':\mathbb{B}^n\to\mathbb{B}^n$ be a diffeomorphism which is identity near the boundary and $F'(0)=q$. Then $(F')^{-1}\circ F$ maps $0$ to $0$. And $$ [0,1]\ni s \to F'(s((F')^{-1}\circ F)(\frac{x}{s})) $$ gives an isotopy between $F$ when $s=1$ and $F'$ when $s=0$. Moreover $p=0$ is mapped to $q$ for all $s\in [0,1]$. This isotopy is a continuous path of diffeomorphisms, but it is not smooth at $s=0$.

3. There are orientation preserving diffeomorphisms of $\mathbb{S}^6$ that are not isotopic to identity map. My understanding is that here we mean a smooth isotopy so the construction from the part 2 does not apply. Indeed, exotic $7$-spheres are obtained by gluing together two balls $\mathbb{B}^7$ along a diffeomorphism $F:\mathbb{S}^6\to\mathbb{S}^6$, see https://en.wikipedia.org/wiki/Exotic_sphere. If $F$ is smoothly isotopic to identity, then we obtain a standard sphere. Note that every orientation preserving diffeomorphism of $\mathbb{S}^6$ is isotopic to a diffeomorphism that is identity on the half sphere because near any point the difeeomorphism is close to a tangent map so this example applies to your question.

Remark. Regarding the construction in part 3, it is not my area of expertise to any comments are welcome.

1. If we do not assume that the diffeomorphism is identity outside a small ball, the answer is no. If $F$ preserves orientation but $F'$ reverses orientation, then there is no isotopy between $F$ and $F'$, but even if both diffeomorphisms preserve orientation there are counterexamples.

Here is one. Take $M=\mathbb{S}^1\times\mathbb{S}^1$ and $F:M\to M$ to be the identity. Let $F':M\to M$, $F'(z_1,z_2)=(z_2,\overline{z}_1)$ be the diffeomorphism that "changes the circles" (we take the complex conjugate - reflection in the $y$-axis to make sure that $F'$ preserves orientation). Fix $p=q=(1,1)\in\mathbb{S}^1\times\mathbb{S}^1$. Then $F(p)=F'(p)=q$. But there is no isotopy that would deform $F'$ to $F$, because $F$ and $F'$ are not homotopic. Indeed, $F(\mathbb{S}^1\times \{1\})=\mathbb{S}^1\times \{1\}$ but $F(\mathbb{S}^1\times \{1\})=\{1\}\times\mathbb{S}^1$ give two different elements in $\pi_1(M)$.

2. The answer is yes if $F$ is identity outside a small ball, but the isotopy is not smooth at the endpoint. We can assume that $F:\mathbb{B}^n\to\mathbb{B}^n$ is identity near the boundary, $p=0$ and $q\in \mathbb{B}^n$. According to our assumptions $F':\mathbb{B}^n\to\mathbb{B}^n$ be a diffeomorphism which is identity near the boundary and $F'(0)=q$. Then $(F')^{-1}\circ F$ maps $0$ to $0$. And $$ [0,1]\ni s \to F'(s((F')^{-1}\circ F)(\frac{x}{s})) $$ gives an isotopy between $F$ when $s=1$ and $F'$ when $s=0$. Moreover $p=0$ is mapped to $q$ for all $s\in [0,1]$. This isotopy is a continuous path of diffeomorphisms, but it is not smooth at $s=0$.

3. There are orientation preserving diffeomorphisms of $\mathbb{S}^6$ that are not isotopic to identity map. My understanding is that here we mean a smooth isotopy so the construction from the part 2 does not apply. Indeed, exotic $7$-spheres are obtained by gluing together two balls $\mathbb{B}^7$ along a diffeomorphism $F:\mathbb{S}^6\to\mathbb{S}^6$, see https://en.wikipedia.org/wiki/Exotic_sphere. If $F$ is smoothly isotopic to identity, then we obtain a standard sphere. Note that every orientation preserving diffeomorphism of $\mathbb{S}^6$ is isotopic to a diffeomorphism that is identity on the half sphere because near any point the difeeomorphism is close to a tangent map so this example applies to your question.

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Piotr Hajlasz
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1. If we do not assume that the diffeomorphism is identity outside a small ball, the answer is no. If $F$ preserves orientation but $F'$ reverses orientation, then there is no isotopy between $F$ and $F'$, but even if both diffeomorphisms preserve orientation there are counterexamples.

Here is one. Take $M=\mathbb{S}^1\times\mathbb{S}^1$ and $F:M\to M$ to be the identity. Let $F':M\to M$, $F'(z_1,z_2)=(z_2,\overline{z}_1)$ be the diffeomorphism that "changes the circles" (we take the complex conjugate - reflection in the $y$-axis to make sure that $F'$ preserves orientation). Fix $p=q=(1,1)\in\mathbb{S}^1\times\mathbb{S}^1$. Then $F(p)=F'(p)=q$. But there is no isotopy that would deform $F'$ to $F$, because $F$ and $F'$ are not homotopic. Indeed, $F(\mathbb{S}^1\times \{1\})=\mathbb{S}^1\times \{1\}$ but $F(\mathbb{S}^1\times \{1\})=\{1\}\times\mathbb{S}^1$ give two different elements in $\pi_1(M)$.

2. The answer is yes if $F$ is identity outside a small ball, but the isotopy is not smooth at the endpoint. We can assume that $F:\mathbb{B}^n\to\mathbb{B}^n$ is identity near the boundary, $p=0$ and $q\in \mathbb{B}^n$. According to our assumptions $F':\mathbb{B}^n\to\mathbb{B}^n$ be a diffeomorphism which is identity near the boundary and $F'(0)=q$. Then $(F')^{-1}\circ F$ maps $0$ to $0$. And $$ [0,1]\ni s \to F'(s((F')^{-1}\circ F)(\frac{x}{s})) $$ gives an isotopy between $F$ when $s=1$ and $F'$ when $s=0$. Moreover $p=0$ is mapped to $q$ for all $s\in [0,1]$. This isotopy is a continuous path of diffeomorphisms, but it is not smooth at $s=0$.

3. There are orientation preserving diffeomorphisms of $\mathbb{S}^6$ that are not isotopic to identity map. My understanding is that here we mean a smooth isotopy so the construction from the part 2 does not apply. Indeed, exotic $7$-spheres are obtained by gluing together two balls $\mathbb{B}^7$ along a diffeomorphism $F:\mathbb{S}^6\to\mathbb{S}^6$, see https://en.wikipedia.org/wiki/Exotic_sphere. If $F$ is smoothly isotopic to identity, then we obtain a standard sphere. Note that every orientation preserving diffeomorphism of $\mathbb{S}^6$ is isotopic to a diffeomorphism that is identity on the half sphere because near any point the difeeomorphism is close to a tangent map so this example applies to your question.

Remark. Regarding the construction in part 3, it is not my area of expertise to any comments are welcome.

1. If we do not assume that the diffeomorphism is identity outside a small ball, the answer is no. If $F$ preserves orientation but $F'$ reverses orientation, then there is no isotopy between $F$ and $F'$, but even if both diffeomorphisms preserve orientation there are counterexamples.

Here is one. Take $M=\mathbb{S}^1\times\mathbb{S}^1$ and $F:M\to M$ to be the identity. Let $F':M\to M$, $F'(z_1,z_2)=(z_2,\overline{z}_1)$ be the diffeomorphism that "changes the circles" (we take the complex conjugate - reflection in the $y$-axis to make sure that $F'$ preserves orientation). Fix $p=q=(1,1)\in\mathbb{S}^1\times\mathbb{S}^1$. Then $F(p)=F'(p)=q$. But there is no isotopy that would deform $F'$ to $F$, because $F$ and $F'$ are not homotopic. Indeed, $F(\mathbb{S}^1\times \{1\})=\mathbb{S}^1\times \{1\}$ but $F(\mathbb{S}^1\times \{1\})=\{1\}\times\mathbb{S}^1$ give two different elements in $\pi_1(M)$.

2. The answer is yes if $F$ is identity outside a small ball, but the isotopy is not smooth at the endpoint. We can assume that $F:\mathbb{B}^n\to\mathbb{B}^n$ is identity near the boundary, $p=0$ and $q\in \mathbb{B}^n$. According to our assumptions $F':\mathbb{B}^n\to\mathbb{B}^n$ be a diffeomorphism which is identity near the boundary and $F'(0)=q$. Then $(F')^{-1}\circ F$ maps $0$ to $0$. And $$ [0,1]\ni s \to F'(s((F')^{-1}\circ F)(\frac{x}{s})) $$ gives an isotopy between $F$ when $s=1$ and $F'$ when $s=0$. Moreover $p=0$ is mapped to $q$ for all $s\in [0,1]$. This isotopy is a continuous path of diffeomorphisms, but it is not smooth at $s=0$.

1. If we do not assume that the diffeomorphism is identity outside a small ball, the answer is no. If $F$ preserves orientation but $F'$ reverses orientation, then there is no isotopy between $F$ and $F'$, but even if both diffeomorphisms preserve orientation there are counterexamples.

Here is one. Take $M=\mathbb{S}^1\times\mathbb{S}^1$ and $F:M\to M$ to be the identity. Let $F':M\to M$, $F'(z_1,z_2)=(z_2,\overline{z}_1)$ be the diffeomorphism that "changes the circles" (we take the complex conjugate - reflection in the $y$-axis to make sure that $F'$ preserves orientation). Fix $p=q=(1,1)\in\mathbb{S}^1\times\mathbb{S}^1$. Then $F(p)=F'(p)=q$. But there is no isotopy that would deform $F'$ to $F$, because $F$ and $F'$ are not homotopic. Indeed, $F(\mathbb{S}^1\times \{1\})=\mathbb{S}^1\times \{1\}$ but $F(\mathbb{S}^1\times \{1\})=\{1\}\times\mathbb{S}^1$ give two different elements in $\pi_1(M)$.

2. The answer is yes if $F$ is identity outside a small ball, but the isotopy is not smooth at the endpoint. We can assume that $F:\mathbb{B}^n\to\mathbb{B}^n$ is identity near the boundary, $p=0$ and $q\in \mathbb{B}^n$. According to our assumptions $F':\mathbb{B}^n\to\mathbb{B}^n$ be a diffeomorphism which is identity near the boundary and $F'(0)=q$. Then $(F')^{-1}\circ F$ maps $0$ to $0$. And $$ [0,1]\ni s \to F'(s((F')^{-1}\circ F)(\frac{x}{s})) $$ gives an isotopy between $F$ when $s=1$ and $F'$ when $s=0$. Moreover $p=0$ is mapped to $q$ for all $s\in [0,1]$. This isotopy is a continuous path of diffeomorphisms, but it is not smooth at $s=0$.

3. There are orientation preserving diffeomorphisms of $\mathbb{S}^6$ that are not isotopic to identity map. My understanding is that here we mean a smooth isotopy so the construction from the part 2 does not apply. Indeed, exotic $7$-spheres are obtained by gluing together two balls $\mathbb{B}^7$ along a diffeomorphism $F:\mathbb{S}^6\to\mathbb{S}^6$, see https://en.wikipedia.org/wiki/Exotic_sphere. If $F$ is smoothly isotopic to identity, then we obtain a standard sphere. Note that every orientation preserving diffeomorphism of $\mathbb{S}^6$ is isotopic to a diffeomorphism that is identity on the half sphere because near any point the difeeomorphism is close to a tangent map so this example applies to your question.

Remark. Regarding the construction in part 3, it is not my area of expertise to any comments are welcome.

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Piotr Hajlasz
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If1. If we do not assume that the diffeomorphism is identity outside a small ball, the answer is no. If $F$ preserves orientation but $F'$ reverses orientation, then there is no isotopy between $F$ and $F'$, but even if both diffeomorphisms preserve orientation there are counterexamples.

Here is one. Take $M=\mathbb{S}^1\times\mathbb{S}^1$ and $F:M\to M$ to be the identity. Let $F':M\to M$, $F'(z_1,z_2)=(z_2,\overline{z}_1)$ be the diffeomorphism that "changes the circles" (we take the complex conjugate - reflection in the $y$-axis to make sure that $F'$ preserves orientation). Fix $p=q=(1,1)\in\mathbb{S}^1\times\mathbb{S}^1$. Then $F(p)=F'(p)=q$. But there is no isotopy that would deform $F'$ to $F$, because $F$ and $F'$ are not homotopic. Indeed, $F(\mathbb{S}^1\times \{1\})=\mathbb{S}^1\times \{1\}$ but $F(\mathbb{S}^1\times \{1\})=\{1\}\times\mathbb{S}^1$ give two different elements in $\pi_1(M)$.

2. The answer is yes if $F$ is identity outside a small ball, but the isotopy is not smooth at the endpoint. We can assume that $F:\mathbb{B}^n\to\mathbb{B}^n$ is identity near the boundary, $p=0$ and $q\in \mathbb{B}^n$. According to our assumptions $F':\mathbb{B}^n\to\mathbb{B}^n$ be a diffeomorphism which is identity near the boundary and $F'(0)=q$. Then $(F')^{-1}\circ F$ maps $0$ to $0$. And $$ [0,1]\ni s \to F'(s((F')^{-1}\circ F)(\frac{x}{s})) $$ gives an isotopy between $F$ when $s=1$ and $F'$ when $s=0$. Moreover $p=0$ is mapped to $q$ for all $s\in [0,1]$. This isotopy is a continuous path of diffeomorphisms, but it is not smooth at $s=0$.

If we do not assume that the diffeomorphism is identity outside a small ball, the answer is no. If $F$ preserves orientation but $F'$ reverses orientation, then there is no isotopy between $F$ and $F'$, but even if both diffeomorphisms preserve orientation there are counterexamples.

Here is one. Take $M=\mathbb{S}^1\times\mathbb{S}^1$ and $F:M\to M$ to be the identity. Let $F':M\to M$, $F'(z_1,z_2)=(z_2,\overline{z}_1)$ be the diffeomorphism that "changes the circles" (we take the complex conjugate - reflection in the $y$-axis to make sure that $F'$ preserves orientation). Fix $p=q=(1,1)\in\mathbb{S}^1\times\mathbb{S}^1$. Then $F(p)=F'(p)=q$. But there is no isotopy that would deform $F'$ to $F$, because $F$ and $F'$ are not homotopic. Indeed, $F(\mathbb{S}^1\times \{1\})=\mathbb{S}^1\times \{1\}$ but $F(\mathbb{S}^1\times \{1\})=\{1\}\times\mathbb{S}^1$ give two different elements in $\pi_1(M)$.

1. If we do not assume that the diffeomorphism is identity outside a small ball, the answer is no. If $F$ preserves orientation but $F'$ reverses orientation, then there is no isotopy between $F$ and $F'$, but even if both diffeomorphisms preserve orientation there are counterexamples.

Here is one. Take $M=\mathbb{S}^1\times\mathbb{S}^1$ and $F:M\to M$ to be the identity. Let $F':M\to M$, $F'(z_1,z_2)=(z_2,\overline{z}_1)$ be the diffeomorphism that "changes the circles" (we take the complex conjugate - reflection in the $y$-axis to make sure that $F'$ preserves orientation). Fix $p=q=(1,1)\in\mathbb{S}^1\times\mathbb{S}^1$. Then $F(p)=F'(p)=q$. But there is no isotopy that would deform $F'$ to $F$, because $F$ and $F'$ are not homotopic. Indeed, $F(\mathbb{S}^1\times \{1\})=\mathbb{S}^1\times \{1\}$ but $F(\mathbb{S}^1\times \{1\})=\{1\}\times\mathbb{S}^1$ give two different elements in $\pi_1(M)$.

2. The answer is yes if $F$ is identity outside a small ball, but the isotopy is not smooth at the endpoint. We can assume that $F:\mathbb{B}^n\to\mathbb{B}^n$ is identity near the boundary, $p=0$ and $q\in \mathbb{B}^n$. According to our assumptions $F':\mathbb{B}^n\to\mathbb{B}^n$ be a diffeomorphism which is identity near the boundary and $F'(0)=q$. Then $(F')^{-1}\circ F$ maps $0$ to $0$. And $$ [0,1]\ni s \to F'(s((F')^{-1}\circ F)(\frac{x}{s})) $$ gives an isotopy between $F$ when $s=1$ and $F'$ when $s=0$. Moreover $p=0$ is mapped to $q$ for all $s\in [0,1]$. This isotopy is a continuous path of diffeomorphisms, but it is not smooth at $s=0$.

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