Proof of Smooth Embeddings and Smooth FunctionsDoes possibility of extension of smooth function impose embedded nature?Embedded Submanifolds Have a Unique Smooth StructureGluing submanifolds along their common boundaryIf $F:Mto N$ is a smooth embedding, then so is $dF:TMto TN$.Smooth coverings are open maps proof verificationManifolds , Submanifolds and EmbeddingsConstructing a submanifold from a submanifold and smooth mapSubset of Open submanifold is a submanifold?Adapted charts for smooth manifoldWhy is an inclusion map from an open subset smooth?
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Proof of Smooth Embeddings and Smooth Functions
Does possibility of extension of smooth function impose embedded nature?Embedded Submanifolds Have a Unique Smooth StructureGluing submanifolds along their common boundaryIf $F:Mto N$ is a smooth embedding, then so is $dF:TMto TN$.Smooth coverings are open maps proof verificationManifolds , Submanifolds and EmbeddingsConstructing a submanifold from a submanifold and smooth mapSubset of Open submanifold is a submanifold?Adapted charts for smooth manifoldWhy is an inclusion map from an open subset smooth?
$begingroup$
Be M a smooth manifold, $N subseteq M$ a submanifold and $ι : N rightarrow M$ the embedding, and be P another smooth manifold. I want to show the following:
(a) If $f:Mrightarrow P$ is smooth ,the restriction $f|_N :Nrightarrow P$ is smooth.
(b) A map $f : P rightarrow N$ is smooth if $ι circ f : P rightarrow M$ is smooth.
For a: If $f:M rightarrow P$ is smooth, there is an open subset $U subseteq M$ with $f_U$ smooth. $Nsubseteq M$ is an embedded manifold, therefore one can choose a neightborhood $U$ such that $U cap N subseteq U$. Hence, there is for every point in $N$ a function $f_U cap N: U cap N rightarrow P$ and because one can chooose an arbitrary large ambient $U$, the function $f_N$ is smooth.
Is this argument correct? And how would I extend it to the second question?
differential-geometry
$endgroup$
add a comment |
$begingroup$
Be M a smooth manifold, $N subseteq M$ a submanifold and $ι : N rightarrow M$ the embedding, and be P another smooth manifold. I want to show the following:
(a) If $f:Mrightarrow P$ is smooth ,the restriction $f|_N :Nrightarrow P$ is smooth.
(b) A map $f : P rightarrow N$ is smooth if $ι circ f : P rightarrow M$ is smooth.
For a: If $f:M rightarrow P$ is smooth, there is an open subset $U subseteq M$ with $f_U$ smooth. $Nsubseteq M$ is an embedded manifold, therefore one can choose a neightborhood $U$ such that $U cap N subseteq U$. Hence, there is for every point in $N$ a function $f_U cap N: U cap N rightarrow P$ and because one can chooose an arbitrary large ambient $U$, the function $f_N$ is smooth.
Is this argument correct? And how would I extend it to the second question?
differential-geometry
$endgroup$
$begingroup$
Here is a more straightforward argument for (a). The embedding $i$ is smooth by definition, and the restriction is just $fcirc i$, which is smooth as the composition of smooth maps.
$endgroup$
– Amitai Yuval
11 hours ago
$begingroup$
Yes, I thought of that argument. But for an exam or smth. would this be enough?
$endgroup$
– John Smith
11 hours ago
add a comment |
$begingroup$
Be M a smooth manifold, $N subseteq M$ a submanifold and $ι : N rightarrow M$ the embedding, and be P another smooth manifold. I want to show the following:
(a) If $f:Mrightarrow P$ is smooth ,the restriction $f|_N :Nrightarrow P$ is smooth.
(b) A map $f : P rightarrow N$ is smooth if $ι circ f : P rightarrow M$ is smooth.
For a: If $f:M rightarrow P$ is smooth, there is an open subset $U subseteq M$ with $f_U$ smooth. $Nsubseteq M$ is an embedded manifold, therefore one can choose a neightborhood $U$ such that $U cap N subseteq U$. Hence, there is for every point in $N$ a function $f_U cap N: U cap N rightarrow P$ and because one can chooose an arbitrary large ambient $U$, the function $f_N$ is smooth.
Is this argument correct? And how would I extend it to the second question?
differential-geometry
$endgroup$
Be M a smooth manifold, $N subseteq M$ a submanifold and $ι : N rightarrow M$ the embedding, and be P another smooth manifold. I want to show the following:
(a) If $f:Mrightarrow P$ is smooth ,the restriction $f|_N :Nrightarrow P$ is smooth.
(b) A map $f : P rightarrow N$ is smooth if $ι circ f : P rightarrow M$ is smooth.
For a: If $f:M rightarrow P$ is smooth, there is an open subset $U subseteq M$ with $f_U$ smooth. $Nsubseteq M$ is an embedded manifold, therefore one can choose a neightborhood $U$ such that $U cap N subseteq U$. Hence, there is for every point in $N$ a function $f_U cap N: U cap N rightarrow P$ and because one can chooose an arbitrary large ambient $U$, the function $f_N$ is smooth.
Is this argument correct? And how would I extend it to the second question?
differential-geometry
differential-geometry
edited 14 hours ago
John Smith
asked 15 hours ago
John SmithJohn Smith
919
919
$begingroup$
Here is a more straightforward argument for (a). The embedding $i$ is smooth by definition, and the restriction is just $fcirc i$, which is smooth as the composition of smooth maps.
$endgroup$
– Amitai Yuval
11 hours ago
$begingroup$
Yes, I thought of that argument. But for an exam or smth. would this be enough?
$endgroup$
– John Smith
11 hours ago
add a comment |
$begingroup$
Here is a more straightforward argument for (a). The embedding $i$ is smooth by definition, and the restriction is just $fcirc i$, which is smooth as the composition of smooth maps.
$endgroup$
– Amitai Yuval
11 hours ago
$begingroup$
Yes, I thought of that argument. But for an exam or smth. would this be enough?
$endgroup$
– John Smith
11 hours ago
$begingroup$
Here is a more straightforward argument for (a). The embedding $i$ is smooth by definition, and the restriction is just $fcirc i$, which is smooth as the composition of smooth maps.
$endgroup$
– Amitai Yuval
11 hours ago
$begingroup$
Here is a more straightforward argument for (a). The embedding $i$ is smooth by definition, and the restriction is just $fcirc i$, which is smooth as the composition of smooth maps.
$endgroup$
– Amitai Yuval
11 hours ago
$begingroup$
Yes, I thought of that argument. But for an exam or smth. would this be enough?
$endgroup$
– John Smith
11 hours ago
$begingroup$
Yes, I thought of that argument. But for an exam or smth. would this be enough?
$endgroup$
– John Smith
11 hours ago
add a comment |
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$begingroup$
Here is a more straightforward argument for (a). The embedding $i$ is smooth by definition, and the restriction is just $fcirc i$, which is smooth as the composition of smooth maps.
$endgroup$
– Amitai Yuval
11 hours ago
$begingroup$
Yes, I thought of that argument. But for an exam or smth. would this be enough?
$endgroup$
– John Smith
11 hours ago