Definition of surjectivity on topological spaces.Can continuity be proven in terms of closed sets?Continuous mapping between topological spacesGeneral Topology: Quotient MapHomeomorphisms of topological spaces, are the topologies isomorphic?What does continuity of a function mapping a topological space to a real line interval mean?A topological space is trivial iff every function to it is continuousopen map from a topological space whose connected components aren't open to a connected spaceMaps between topological spaces; any rules?Question about the difference between topological spaces and topologiesf nonsurjective map between two topological spaces; continuity
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Definition of surjectivity on topological spaces.
Can continuity be proven in terms of closed sets?Continuous mapping between topological spacesGeneral Topology: Quotient MapHomeomorphisms of topological spaces, are the topologies isomorphic?What does continuity of a function mapping a topological space to a real line interval mean?A topological space is trivial iff every function to it is continuousopen map from a topological space whose connected components aren't open to a connected spaceMaps between topological spaces; any rules?Question about the difference between topological spaces and topologiesf nonsurjective map between two topological spaces; continuity
$begingroup$
If a map $f$ between topological spaces, $X$ and $Y$, is surjective, does that mean that every open set in the topological space $Y$ has an open preimage, or merely that the set $Y$ has an open preimage? I guess my question is: is surjectivity mean hitting all the the elements of the set $Y$, or all the open sets on the topological space $Y$? Does it depend on the context?
Thanks 😃
general-topology
asked Mar 22 at 0:40
PatrickPatrick
1038
$endgroup$
add a comment |
$begingroup$
If a map $f$ between topological spaces, $X$ and $Y$, is surjective, does that mean that every open set in the topological space $Y$ has an open preimage, or merely that the set $Y$ has an open preimage? I guess my question is: is surjectivity mean hitting all the the elements of the set $Y$, or all the open sets on the topological space $Y$? Does it depend on the context?
Thanks 😃
general-topology
asked Mar 22 at 0:40
PatrickPatrick
1038
$endgroup$
1
$begingroup$
Technically both are true- surjectivity means that each element in $Y$ is hit at least once so all open sets are also 'hit' (by subsets of $X$) but the preimage of each open set in $Y$ will always be open in $X$ iff $f$ is continuous
$endgroup$
– Cardioid_Ass_22
Mar 22 at 0:43
$begingroup$
@Cardioid_Ass_22 that clears everything up. Thank you :)
$endgroup$
– Patrick
Mar 22 at 0:50
add a comment |
$begingroup$
If a map $f$ between topological spaces, $X$ and $Y$, is surjective, does that mean that every open set in the topological space $Y$ has an open preimage, or merely that the set $Y$ has an open preimage? I guess my question is: is surjectivity mean hitting all the the elements of the set $Y$, or all the open sets on the topological space $Y$? Does it depend on the context?
Thanks 😃
general-topology
asked Mar 22 at 0:40
PatrickPatrick
1038
$endgroup$
If a map $f$ between topological spaces, $X$ and $Y$, is surjective, does that mean that every open set in the topological space $Y$ has an open preimage, or merely that the set $Y$ has an open preimage? I guess my question is: is surjectivity mean hitting all the the elements of the set $Y$, or all the open sets on the topological space $Y$? Does it depend on the context?
Thanks 😃
general-topology
general-topology
asked Mar 22 at 0:40
PatrickPatrick
1038
asked Mar 22 at 0:40
PatrickPatrick
1038
asked Mar 22 at 0:40
PatrickPatrick
1038
asked Mar 22 at 0:40
PatrickPatrick
1038
asked Mar 22 at 0:40
PatrickPatrick
1038
1038
1
$begingroup$
Technically both are true- surjectivity means that each element in $Y$ is hit at least once so all open sets are also 'hit' (by subsets of $X$) but the preimage of each open set in $Y$ will always be open in $X$ iff $f$ is continuous
$endgroup$
– Cardioid_Ass_22
Mar 22 at 0:43
$begingroup$
@Cardioid_Ass_22 that clears everything up. Thank you :)
$endgroup$
– Patrick
Mar 22 at 0:50
add a comment |
1
$begingroup$
Technically both are true- surjectivity means that each element in $Y$ is hit at least once so all open sets are also 'hit' (by subsets of $X$) but the preimage of each open set in $Y$ will always be open in $X$ iff $f$ is continuous
$endgroup$
– Cardioid_Ass_22
Mar 22 at 0:43
$begingroup$
@Cardioid_Ass_22 that clears everything up. Thank you :)
$endgroup$
– Patrick
Mar 22 at 0:50
1
1
$begingroup$
Technically both are true- surjectivity means that each element in $Y$ is hit at least once so all open sets are also 'hit' (by subsets of $X$) but the preimage of each open set in $Y$ will always be open in $X$ iff $f$ is continuous
$endgroup$
– Cardioid_Ass_22
Mar 22 at 0:43
$begingroup$
Technically both are true- surjectivity means that each element in $Y$ is hit at least once so all open sets are also 'hit' (by subsets of $X$) but the preimage of each open set in $Y$ will always be open in $X$ iff $f$ is continuous
$endgroup$
– Cardioid_Ass_22
Mar 22 at 0:43
$begingroup$
@Cardioid_Ass_22 that clears everything up. Thank you :)
$endgroup$
– Patrick
Mar 22 at 0:50
$begingroup$
@Cardioid_Ass_22 that clears everything up. Thank you :)
$endgroup$
– Patrick
Mar 22 at 0:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Unless I'm missing some non-standard definition, surjectivity is not a topological property. Whether a map between two topological spaces $X$ and $Y$ is surjective does not depend one jot on the topology of $X$ or the topology of $Y$.
To say $f : X to Y$ is surjective is to say that, for every $y in Y$, there exists an $x in X$ such that $f(x) = y$. It may not be the case that open subsets of $Y$ have open preimages (i.e. $f$ need not be continuous). It will, however, always be the case that $f^-1(Y) = X$, which is open, as every element of $X$ maps into $Y$, by definition of a function. This is true even when $f$ is not surjective!
answered Mar 22 at 1:03
Theo BenditTheo Bendit
20.3k12353
$endgroup$
add a comment |
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$begingroup$
Unless I'm missing some non-standard definition, surjectivity is not a topological property. Whether a map between two topological spaces $X$ and $Y$ is surjective does not depend one jot on the topology of $X$ or the topology of $Y$.
To say $f : X to Y$ is surjective is to say that, for every $y in Y$, there exists an $x in X$ such that $f(x) = y$. It may not be the case that open subsets of $Y$ have open preimages (i.e. $f$ need not be continuous). It will, however, always be the case that $f^-1(Y) = X$, which is open, as every element of $X$ maps into $Y$, by definition of a function. This is true even when $f$ is not surjective!
answered Mar 22 at 1:03
Theo BenditTheo Bendit
20.3k12353
$endgroup$
add a comment |
$begingroup$
Unless I'm missing some non-standard definition, surjectivity is not a topological property. Whether a map between two topological spaces $X$ and $Y$ is surjective does not depend one jot on the topology of $X$ or the topology of $Y$.
To say $f : X to Y$ is surjective is to say that, for every $y in Y$, there exists an $x in X$ such that $f(x) = y$. It may not be the case that open subsets of $Y$ have open preimages (i.e. $f$ need not be continuous). It will, however, always be the case that $f^-1(Y) = X$, which is open, as every element of $X$ maps into $Y$, by definition of a function. This is true even when $f$ is not surjective!
answered Mar 22 at 1:03
Theo BenditTheo Bendit
20.3k12353
$endgroup$
add a comment |
$begingroup$
Unless I'm missing some non-standard definition, surjectivity is not a topological property. Whether a map between two topological spaces $X$ and $Y$ is surjective does not depend one jot on the topology of $X$ or the topology of $Y$.
To say $f : X to Y$ is surjective is to say that, for every $y in Y$, there exists an $x in X$ such that $f(x) = y$. It may not be the case that open subsets of $Y$ have open preimages (i.e. $f$ need not be continuous). It will, however, always be the case that $f^-1(Y) = X$, which is open, as every element of $X$ maps into $Y$, by definition of a function. This is true even when $f$ is not surjective!
answered Mar 22 at 1:03
Theo BenditTheo Bendit
20.3k12353
$endgroup$
Unless I'm missing some non-standard definition, surjectivity is not a topological property. Whether a map between two topological spaces $X$ and $Y$ is surjective does not depend one jot on the topology of $X$ or the topology of $Y$.
To say $f : X to Y$ is surjective is to say that, for every $y in Y$, there exists an $x in X$ such that $f(x) = y$. It may not be the case that open subsets of $Y$ have open preimages (i.e. $f$ need not be continuous). It will, however, always be the case that $f^-1(Y) = X$, which is open, as every element of $X$ maps into $Y$, by definition of a function. This is true even when $f$ is not surjective!
answered Mar 22 at 1:03
Theo BenditTheo Bendit
20.3k12353
edited Mar 22 at 1:19
answered Mar 22 at 1:03
Theo BenditTheo Bendit
20.3k12353
answered Mar 22 at 1:03
Theo BenditTheo Bendit
20.3k12353
answered Mar 22 at 1:03
Theo BenditTheo Bendit
20.3k12353
20.3k12353
add a comment |
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$begingroup$
Technically both are true- surjectivity means that each element in $Y$ is hit at least once so all open sets are also 'hit' (by subsets of $X$) but the preimage of each open set in $Y$ will always be open in $X$ iff $f$ is continuous
$endgroup$
– Cardioid_Ass_22
Mar 22 at 0:43
$begingroup$
@Cardioid_Ass_22 that clears everything up. Thank you :)
$endgroup$
– Patrick
Mar 22 at 0:50