Definition of a smooth function on $[0,1]^k$
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Let $M$ be a smooth manifold, and $f: [0,1]^k rightarrow M$ be a continuous function which is smooth on $(0,1)^k$. I have seen two definitions of what it means for $f$ to be smooth. First, there exists an open neighborhood $U subset mathbb R^k$ such that $f$ extends to a smooth function on $U$. Second, at each point $p$ on the boundary of $[0,1]^k$, there exists an open neighborhood $U$ of $p$ such that $f$ extends to a smooth function on $U cap (0,1)^k$.
Are these two definitions equivalent? If not, what should be the "correct" definition? Say, for the purpose of defining singular homology and integration of differential forms?
calculus differential-geometry smooth-manifolds
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Let $M$ be a smooth manifold, and $f: [0,1]^k rightarrow M$ be a continuous function which is smooth on $(0,1)^k$. I have seen two definitions of what it means for $f$ to be smooth. First, there exists an open neighborhood $U subset mathbb R^k$ such that $f$ extends to a smooth function on $U$. Second, at each point $p$ on the boundary of $[0,1]^k$, there exists an open neighborhood $U$ of $p$ such that $f$ extends to a smooth function on $U cap (0,1)^k$.
Are these two definitions equivalent? If not, what should be the "correct" definition? Say, for the purpose of defining singular homology and integration of differential forms?
calculus differential-geometry smooth-manifolds
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $M$ be a smooth manifold, and $f: [0,1]^k rightarrow M$ be a continuous function which is smooth on $(0,1)^k$. I have seen two definitions of what it means for $f$ to be smooth. First, there exists an open neighborhood $U subset mathbb R^k$ such that $f$ extends to a smooth function on $U$. Second, at each point $p$ on the boundary of $[0,1]^k$, there exists an open neighborhood $U$ of $p$ such that $f$ extends to a smooth function on $U cap (0,1)^k$.
Are these two definitions equivalent? If not, what should be the "correct" definition? Say, for the purpose of defining singular homology and integration of differential forms?
calculus differential-geometry smooth-manifolds
Let $M$ be a smooth manifold, and $f: [0,1]^k rightarrow M$ be a continuous function which is smooth on $(0,1)^k$. I have seen two definitions of what it means for $f$ to be smooth. First, there exists an open neighborhood $U subset mathbb R^k$ such that $f$ extends to a smooth function on $U$. Second, at each point $p$ on the boundary of $[0,1]^k$, there exists an open neighborhood $U$ of $p$ such that $f$ extends to a smooth function on $U cap (0,1)^k$.
Are these two definitions equivalent? If not, what should be the "correct" definition? Say, for the purpose of defining singular homology and integration of differential forms?
calculus differential-geometry smooth-manifolds
calculus differential-geometry smooth-manifolds
edited Nov 21 at 1:20
asked Nov 21 at 1:04
D_S
13.2k51551
13.2k51551
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2 Answers
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Yes, these definitions are equivalents. The first one clearly implies the second. Using the notion of partition of unity (see here), you can show that the second one also implies the first.
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2
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Yes, both definition are equivalent. You can use partitions of unity: Suppose that for every $x$ in $[0,1]^k$ there exists an open neighborhood of $x$ and a smooth function $f_x:Ucap (0,1)^k to M$ wich agrees with $f$. By compactness of $[0,1]^k$ there exists a finite cover ${U_1,ldots, U_r}$ of $[0,1]^k$ consisting in some of such neighborhoods. Take a partition of unity ${varphi_1,ldots,varphi_r}$ subordinated to the cover and define $F:cup_iU_ito M$ by the formula $F(x)=sum_ivarphi_i(x)f_i(x)$.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yes, these definitions are equivalents. The first one clearly implies the second. Using the notion of partition of unity (see here), you can show that the second one also implies the first.
add a comment |
up vote
1
down vote
accepted
Yes, these definitions are equivalents. The first one clearly implies the second. Using the notion of partition of unity (see here), you can show that the second one also implies the first.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yes, these definitions are equivalents. The first one clearly implies the second. Using the notion of partition of unity (see here), you can show that the second one also implies the first.
Yes, these definitions are equivalents. The first one clearly implies the second. Using the notion of partition of unity (see here), you can show that the second one also implies the first.
answered Nov 21 at 1:18
rldias
2,9851522
2,9851522
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up vote
2
down vote
Yes, both definition are equivalent. You can use partitions of unity: Suppose that for every $x$ in $[0,1]^k$ there exists an open neighborhood of $x$ and a smooth function $f_x:Ucap (0,1)^k to M$ wich agrees with $f$. By compactness of $[0,1]^k$ there exists a finite cover ${U_1,ldots, U_r}$ of $[0,1]^k$ consisting in some of such neighborhoods. Take a partition of unity ${varphi_1,ldots,varphi_r}$ subordinated to the cover and define $F:cup_iU_ito M$ by the formula $F(x)=sum_ivarphi_i(x)f_i(x)$.
add a comment |
up vote
2
down vote
Yes, both definition are equivalent. You can use partitions of unity: Suppose that for every $x$ in $[0,1]^k$ there exists an open neighborhood of $x$ and a smooth function $f_x:Ucap (0,1)^k to M$ wich agrees with $f$. By compactness of $[0,1]^k$ there exists a finite cover ${U_1,ldots, U_r}$ of $[0,1]^k$ consisting in some of such neighborhoods. Take a partition of unity ${varphi_1,ldots,varphi_r}$ subordinated to the cover and define $F:cup_iU_ito M$ by the formula $F(x)=sum_ivarphi_i(x)f_i(x)$.
add a comment |
up vote
2
down vote
up vote
2
down vote
Yes, both definition are equivalent. You can use partitions of unity: Suppose that for every $x$ in $[0,1]^k$ there exists an open neighborhood of $x$ and a smooth function $f_x:Ucap (0,1)^k to M$ wich agrees with $f$. By compactness of $[0,1]^k$ there exists a finite cover ${U_1,ldots, U_r}$ of $[0,1]^k$ consisting in some of such neighborhoods. Take a partition of unity ${varphi_1,ldots,varphi_r}$ subordinated to the cover and define $F:cup_iU_ito M$ by the formula $F(x)=sum_ivarphi_i(x)f_i(x)$.
Yes, both definition are equivalent. You can use partitions of unity: Suppose that for every $x$ in $[0,1]^k$ there exists an open neighborhood of $x$ and a smooth function $f_x:Ucap (0,1)^k to M$ wich agrees with $f$. By compactness of $[0,1]^k$ there exists a finite cover ${U_1,ldots, U_r}$ of $[0,1]^k$ consisting in some of such neighborhoods. Take a partition of unity ${varphi_1,ldots,varphi_r}$ subordinated to the cover and define $F:cup_iU_ito M$ by the formula $F(x)=sum_ivarphi_i(x)f_i(x)$.
answered Nov 21 at 1:24
Dante Grevino
7787
7787
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