Mixing
Covering space of disconnected spaces
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Let $f:Mto N$ is a $d-$fold covering. Is it true that $M$ is disconnected if $N$ is disconnected? If not, can anyone give a counter example?
covering-spaces
asked Jul 30 at 7:56
usr1988
153
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up vote
1
down vote
favorite
Let $f:Mto N$ is a $d-$fold covering. Is it true that $M$ is disconnected if $N$ is disconnected? If not, can anyone give a counter example?
covering-spaces
asked Jul 30 at 7:56
usr1988
153
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f:Mto N$ is a $d-$fold covering. Is it true that $M$ is disconnected if $N$ is disconnected? If not, can anyone give a counter example?
covering-spaces
asked Jul 30 at 7:56
usr1988
153
Let $f:Mto N$ is a $d-$fold covering. Is it true that $M$ is disconnected if $N$ is disconnected? If not, can anyone give a counter example?
covering-spaces
asked Jul 30 at 7:56
usr1988
153
asked Jul 30 at 7:56
usr1988
153
asked Jul 30 at 7:56
usr1988
153
asked Jul 30 at 7:56
usr1988
153
153
add a comment |Â
add a comment |Â
1 Answer
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oldest
votes
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0
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Yes, if $N$ is disconnected so is $M$. This is a general case of the result "Let $f :M to N$ be a continuous surjection, and $N$ disconnected. Then $M$ is disconnected."
To prove this, decompose $N$ as $N=Ucup V$ for non-empty disjoint open sets $U,V$.
Then $M=f^-1(U) cup f^-1(V)$ and $f^-1(U), f^-1(V)$ are both
- non-empty (since $f$ is a surjection)
- disjoint (because $U$ and $V$ are)
- open (because $U$ and $V$ are, and $f$ is continuous).
answered Jul 30 at 8:12
Daniel Mroz
851314
Thanks for your quick answer. But I still do not understand why $f^-1(U)$ and $f^-1(V)$ are disjoint. Could you please shed light on this?
– usr1988
Jul 30 at 8:25
Suppose they are not, so there exists some $x in f^-1(U) cap f^-1(V)$. This means that $f(x) in U$ and $f(x) in V$. But that isn't possible, since $U, V$ are disjoint.
– Daniel Mroz
Jul 30 at 8:28
Oh, yes. Thank you again.
– usr1988
Jul 30 at 8:30
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Yes, if $N$ is disconnected so is $M$. This is a general case of the result "Let $f :M to N$ be a continuous surjection, and $N$ disconnected. Then $M$ is disconnected."
To prove this, decompose $N$ as $N=Ucup V$ for non-empty disjoint open sets $U,V$.
Then $M=f^-1(U) cup f^-1(V)$ and $f^-1(U), f^-1(V)$ are both
- non-empty (since $f$ is a surjection)
- disjoint (because $U$ and $V$ are)
- open (because $U$ and $V$ are, and $f$ is continuous).
answered Jul 30 at 8:12
Daniel Mroz
851314
Thanks for your quick answer. But I still do not understand why $f^-1(U)$ and $f^-1(V)$ are disjoint. Could you please shed light on this?
– usr1988
Jul 30 at 8:25
Suppose they are not, so there exists some $x in f^-1(U) cap f^-1(V)$. This means that $f(x) in U$ and $f(x) in V$. But that isn't possible, since $U, V$ are disjoint.
– Daniel Mroz
Jul 30 at 8:28
Oh, yes. Thank you again.
– usr1988
Jul 30 at 8:30
add a comment |Â
up vote
0
down vote
accepted
Yes, if $N$ is disconnected so is $M$. This is a general case of the result "Let $f :M to N$ be a continuous surjection, and $N$ disconnected. Then $M$ is disconnected."
To prove this, decompose $N$ as $N=Ucup V$ for non-empty disjoint open sets $U,V$.
Then $M=f^-1(U) cup f^-1(V)$ and $f^-1(U), f^-1(V)$ are both
- non-empty (since $f$ is a surjection)
- disjoint (because $U$ and $V$ are)
- open (because $U$ and $V$ are, and $f$ is continuous).
answered Jul 30 at 8:12
Daniel Mroz
851314
Thanks for your quick answer. But I still do not understand why $f^-1(U)$ and $f^-1(V)$ are disjoint. Could you please shed light on this?
– usr1988
Jul 30 at 8:25
Suppose they are not, so there exists some $x in f^-1(U) cap f^-1(V)$. This means that $f(x) in U$ and $f(x) in V$. But that isn't possible, since $U, V$ are disjoint.
– Daniel Mroz
Jul 30 at 8:28
Oh, yes. Thank you again.
– usr1988
Jul 30 at 8:30
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Yes, if $N$ is disconnected so is $M$. This is a general case of the result "Let $f :M to N$ be a continuous surjection, and $N$ disconnected. Then $M$ is disconnected."
To prove this, decompose $N$ as $N=Ucup V$ for non-empty disjoint open sets $U,V$.
Then $M=f^-1(U) cup f^-1(V)$ and $f^-1(U), f^-1(V)$ are both
- non-empty (since $f$ is a surjection)
- disjoint (because $U$ and $V$ are)
- open (because $U$ and $V$ are, and $f$ is continuous).
answered Jul 30 at 8:12
Daniel Mroz
851314
Yes, if $N$ is disconnected so is $M$. This is a general case of the result "Let $f :M to N$ be a continuous surjection, and $N$ disconnected. Then $M$ is disconnected."
To prove this, decompose $N$ as $N=Ucup V$ for non-empty disjoint open sets $U,V$.
Then $M=f^-1(U) cup f^-1(V)$ and $f^-1(U), f^-1(V)$ are both
- non-empty (since $f$ is a surjection)
- disjoint (because $U$ and $V$ are)
- open (because $U$ and $V$ are, and $f$ is continuous).
answered Jul 30 at 8:12
Daniel Mroz
851314
answered Jul 30 at 8:12
Daniel Mroz
851314
answered Jul 30 at 8:12
Daniel Mroz
851314
answered Jul 30 at 8:12
Daniel Mroz
851314
851314
Thanks for your quick answer. But I still do not understand why $f^-1(U)$ and $f^-1(V)$ are disjoint. Could you please shed light on this?
– usr1988
Jul 30 at 8:25
Suppose they are not, so there exists some $x in f^-1(U) cap f^-1(V)$. This means that $f(x) in U$ and $f(x) in V$. But that isn't possible, since $U, V$ are disjoint.
– Daniel Mroz
Jul 30 at 8:28
Oh, yes. Thank you again.
– usr1988
Jul 30 at 8:30
add a comment |Â
Thanks for your quick answer. But I still do not understand why $f^-1(U)$ and $f^-1(V)$ are disjoint. Could you please shed light on this?
– usr1988
Jul 30 at 8:25
Suppose they are not, so there exists some $x in f^-1(U) cap f^-1(V)$. This means that $f(x) in U$ and $f(x) in V$. But that isn't possible, since $U, V$ are disjoint.
– Daniel Mroz
Jul 30 at 8:28
Oh, yes. Thank you again.
– usr1988
Jul 30 at 8:30
Thanks for your quick answer. But I still do not understand why $f^-1(U)$ and $f^-1(V)$ are disjoint. Could you please shed light on this?
– usr1988
Jul 30 at 8:25
Thanks for your quick answer. But I still do not understand why $f^-1(U)$ and $f^-1(V)$ are disjoint. Could you please shed light on this?
– usr1988
Jul 30 at 8:25
Suppose they are not, so there exists some $x in f^-1(U) cap f^-1(V)$. This means that $f(x) in U$ and $f(x) in V$. But that isn't possible, since $U, V$ are disjoint.
– Daniel Mroz
Jul 30 at 8:28
Suppose they are not, so there exists some $x in f^-1(U) cap f^-1(V)$. This means that $f(x) in U$ and $f(x) in V$. But that isn't possible, since $U, V$ are disjoint.
– Daniel Mroz
Jul 30 at 8:28
Oh, yes. Thank you again.
– usr1988
Jul 30 at 8:30
Oh, yes. Thank you again.
– usr1988
Jul 30 at 8:30
add a comment |Â
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