f nonsurjective map between two topological spaces; continuity
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Simple question: Suppose $f:Xto Y$ is a nonsurjective map between X,Y topological spaces where Y has the discrete topology. Then there is an $yin Y$ such that its preimage does not exist; i.e. $f^{-1}(y)$ does not exist. Since Y has the discrete topology, y is closed in Y. Suppose we want f to be continuous, then we want that the preimage of every closed set of Y to be closed in X. But although y is closed, $f^{-1}(y)$ does not exist. Then f is never continuous in this case?
In general, for $f:Xto Y$ nonsurjective map between X,Y topological spaces, how do we think about preimage of open sets of Y when some elements of X are not mapped to elements contained in a open set of Y?
general-topology continuity
edited Jan 14 at 8:31
José Carlos Santos
169k23132237
asked Jan 14 at 8:27
metalder9metalder9
647
$endgroup$
add a comment |
$begingroup$
Simple question: Suppose $f:Xto Y$ is a nonsurjective map between X,Y topological spaces where Y has the discrete topology. Then there is an $yin Y$ such that its preimage does not exist; i.e. $f^{-1}(y)$ does not exist. Since Y has the discrete topology, y is closed in Y. Suppose we want f to be continuous, then we want that the preimage of every closed set of Y to be closed in X. But although y is closed, $f^{-1}(y)$ does not exist. Then f is never continuous in this case?
In general, for $f:Xto Y$ nonsurjective map between X,Y topological spaces, how do we think about preimage of open sets of Y when some elements of X are not mapped to elements contained in a open set of Y?
general-topology continuity
edited Jan 14 at 8:31
José Carlos Santos
169k23132237
asked Jan 14 at 8:27
metalder9metalder9
647
$endgroup$
$begingroup$
$f^{-1}(y)$ always exists. It is the set of all $x in X$ such that $f(x) = y$. The only thing is that $f^{-1}(y) = emptyset$ if $y $ is not in the image f $X$.
$endgroup$
– Paul Frost
Jan 14 at 14:59
$begingroup$
@PaulFrost So general question, suppose y is not in the range of $f$ but $f(a)=f(b)=f(c)=z$, is the preimage $f^{−1}({y,z})={a,b,c}$?
$endgroup$
– metalder9
Jan 14 at 17:56
1
$begingroup$
Correct! For any $B subset Y$ the defnition is $f^{-1}(B) = { x in X mid f(x) in B }$.
$endgroup$
– Paul Frost
Jan 14 at 18:00
add a comment |
$begingroup$
Simple question: Suppose $f:Xto Y$ is a nonsurjective map between X,Y topological spaces where Y has the discrete topology. Then there is an $yin Y$ such that its preimage does not exist; i.e. $f^{-1}(y)$ does not exist. Since Y has the discrete topology, y is closed in Y. Suppose we want f to be continuous, then we want that the preimage of every closed set of Y to be closed in X. But although y is closed, $f^{-1}(y)$ does not exist. Then f is never continuous in this case?
In general, for $f:Xto Y$ nonsurjective map between X,Y topological spaces, how do we think about preimage of open sets of Y when some elements of X are not mapped to elements contained in a open set of Y?
general-topology continuity
edited Jan 14 at 8:31
José Carlos Santos
169k23132237
asked Jan 14 at 8:27
metalder9metalder9
647
$endgroup$
Simple question: Suppose $f:Xto Y$ is a nonsurjective map between X,Y topological spaces where Y has the discrete topology. Then there is an $yin Y$ such that its preimage does not exist; i.e. $f^{-1}(y)$ does not exist. Since Y has the discrete topology, y is closed in Y. Suppose we want f to be continuous, then we want that the preimage of every closed set of Y to be closed in X. But although y is closed, $f^{-1}(y)$ does not exist. Then f is never continuous in this case?
In general, for $f:Xto Y$ nonsurjective map between X,Y topological spaces, how do we think about preimage of open sets of Y when some elements of X are not mapped to elements contained in a open set of Y?
general-topology continuity
general-topology continuity
edited Jan 14 at 8:31
José Carlos Santos
169k23132237
asked Jan 14 at 8:27
metalder9metalder9
647
edited Jan 14 at 8:31
José Carlos Santos
169k23132237
asked Jan 14 at 8:27
metalder9metalder9
647
edited Jan 14 at 8:31
José Carlos Santos
169k23132237
edited Jan 14 at 8:31
José Carlos Santos
169k23132237
edited Jan 14 at 8:31
José Carlos Santos
169k23132237
169k23132237
asked Jan 14 at 8:27
metalder9metalder9
647
asked Jan 14 at 8:27
metalder9metalder9
647
asked Jan 14 at 8:27
metalder9metalder9
647
647
$begingroup$
$f^{-1}(y)$ always exists. It is the set of all $x in X$ such that $f(x) = y$. The only thing is that $f^{-1}(y) = emptyset$ if $y $ is not in the image f $X$.
$endgroup$
– Paul Frost
Jan 14 at 14:59
$begingroup$
@PaulFrost So general question, suppose y is not in the range of $f$ but $f(a)=f(b)=f(c)=z$, is the preimage $f^{−1}({y,z})={a,b,c}$?
$endgroup$
– metalder9
Jan 14 at 17:56
1
$begingroup$
Correct! For any $B subset Y$ the defnition is $f^{-1}(B) = { x in X mid f(x) in B }$.
$endgroup$
– Paul Frost
Jan 14 at 18:00
add a comment |
$begingroup$
$f^{-1}(y)$ always exists. It is the set of all $x in X$ such that $f(x) = y$. The only thing is that $f^{-1}(y) = emptyset$ if $y $ is not in the image f $X$.
$endgroup$
– Paul Frost
Jan 14 at 14:59
$begingroup$
@PaulFrost So general question, suppose y is not in the range of $f$ but $f(a)=f(b)=f(c)=z$, is the preimage $f^{−1}({y,z})={a,b,c}$?
$endgroup$
– metalder9
Jan 14 at 17:56
1
$begingroup$
Correct! For any $B subset Y$ the defnition is $f^{-1}(B) = { x in X mid f(x) in B }$.
$endgroup$
– Paul Frost
Jan 14 at 18:00
$begingroup$
$f^{-1}(y)$ always exists. It is the set of all $x in X$ such that $f(x) = y$. The only thing is that $f^{-1}(y) = emptyset$ if $y $ is not in the image f $X$.
$endgroup$
– Paul Frost
Jan 14 at 14:59
$begingroup$
$f^{-1}(y)$ always exists. It is the set of all $x in X$ such that $f(x) = y$. The only thing is that $f^{-1}(y) = emptyset$ if $y $ is not in the image f $X$.
$endgroup$
– Paul Frost
Jan 14 at 14:59
$begingroup$
@PaulFrost So general question, suppose y is not in the range of $f$ but $f(a)=f(b)=f(c)=z$, is the preimage $f^{−1}({y,z})={a,b,c}$?
$endgroup$
– metalder9
Jan 14 at 17:56
$begingroup$
@PaulFrost So general question, suppose y is not in the range of $f$ but $f(a)=f(b)=f(c)=z$, is the preimage $f^{−1}({y,z})={a,b,c}$?
$endgroup$
– metalder9
Jan 14 at 17:56
1
1
$begingroup$
Correct! For any $B subset Y$ the defnition is $f^{-1}(B) = { x in X mid f(x) in B }$.
$endgroup$
– Paul Frost
Jan 14 at 18:00
$begingroup$
Correct! For any $B subset Y$ the defnition is $f^{-1}(B) = { x in X mid f(x) in B }$.
$endgroup$
– Paul Frost
Jan 14 at 18:00
add a comment |
1 Answer
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If $y$ is not in the range of $f$ then $f^{-1}(y)$ does exist. It is the empty set and empty set is open in any topology.
answered Jan 14 at 8:29
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
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$begingroup$
If $y$ is not in the range of $f$ then $f^{-1}(y)$ does exist. It is the empty set and empty set is open in any topology.
answered Jan 14 at 8:29
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
$endgroup$
add a comment |
$begingroup$
If $y$ is not in the range of $f$ then $f^{-1}(y)$ does exist. It is the empty set and empty set is open in any topology.
answered Jan 14 at 8:29
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
$endgroup$
add a comment |
$begingroup$
If $y$ is not in the range of $f$ then $f^{-1}(y)$ does exist. It is the empty set and empty set is open in any topology.
answered Jan 14 at 8:29
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
$endgroup$
If $y$ is not in the range of $f$ then $f^{-1}(y)$ does exist. It is the empty set and empty set is open in any topology.
answered Jan 14 at 8:29
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
answered Jan 14 at 8:29
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
answered Jan 14 at 8:29
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
answered Jan 14 at 8:29
Kavi Rama MurthyKavi Rama Murthy
69.1k53169
69.1k53169
add a comment |
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$begingroup$
$f^{-1}(y)$ always exists. It is the set of all $x in X$ such that $f(x) = y$. The only thing is that $f^{-1}(y) = emptyset$ if $y $ is not in the image f $X$.
$endgroup$
– Paul Frost
Jan 14 at 14:59
$begingroup$
@PaulFrost So general question, suppose y is not in the range of $f$ but $f(a)=f(b)=f(c)=z$, is the preimage $f^{−1}({y,z})={a,b,c}$?
$endgroup$
– metalder9
Jan 14 at 17:56
1
$begingroup$
Correct! For any $B subset Y$ the defnition is $f^{-1}(B) = { x in X mid f(x) in B }$.
$endgroup$
– Paul Frost
Jan 14 at 18:00