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A question regarding the property of a quotient map.
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Let $X$ and $Y$ be two topological spaces. Suppose $f:X longrightarrow Y$ be a quotient map. Consider the equivalence class of $X$ formed by the disjoint non-empty fibres of $f$. Let $X/sim_f$ be the set of all these non-empty disjoint classes of $X$. Consider the map $tilde f : X/sim_f longrightarrow Y$ defined by $tilde f ([x]) = f(x),$ $x in X$. Then this map is well-defined and one-to-one by the construction and also since $f$ is onto, $tilde f$ is onto. Let $p : X longrightarrow X/sim_f$ be the quotient map where $X/sim_f$ is endowed with quotient topology. Then it is not hard to see that $tilde f circ p = f$. SInce $f$ is continuous so by Universal Mapping Property we can say that $tilde f$ is continuous.
Now my question is can we say that $tilde f$ is a homeomorphism? My instructor said that it is true and we were given it to prove.
What I have tried is as follows $:$
In order to show that $tilde f$ is a homeomorphism we need only to prove that it is open. So I take an open subset $tilde V subset X/sim_f$. We need to prove that $tilde f (tilde V)$ is open in $Y$. Now what is $tilde f (tilde V)$? I have found that $tilde f (tilde V) = f(S)$. Where $$S = cup_A in tilde V A.$$ Now since $tilde V$ is open in $X/sim_f$ so $p^-1 ( tilde V )$ is open in $X$. Now what is $p^-1 (tildeV)$? I have found that $p^-1 (tildeV) = S$. So we have to prove that $f(S)$ is an open subset of $Y$, where $S$ is an open subset of $X$.
I tried to prove it by contradiction. If not let $f(S)$ is not an open subset of $Y$. But that implies $f^-1(f(S))$ is not an open subset of $X$. If $f$ is one-to-one we are clearly done since then $f^-1(f(S))=S$ which leads us to a contradiction because $S$ was an open subset of $X$. But if $f$ is not one-to-one I don't know how to proceed. Please help me in this regard.
Thank you very much.
general-topology quotient-spaces
asked 13 hours ago
Debabrata Chattopadhyay.
8311
add a comment |Â
up vote
0
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Let $X$ and $Y$ be two topological spaces. Suppose $f:X longrightarrow Y$ be a quotient map. Consider the equivalence class of $X$ formed by the disjoint non-empty fibres of $f$. Let $X/sim_f$ be the set of all these non-empty disjoint classes of $X$. Consider the map $tilde f : X/sim_f longrightarrow Y$ defined by $tilde f ([x]) = f(x),$ $x in X$. Then this map is well-defined and one-to-one by the construction and also since $f$ is onto, $tilde f$ is onto. Let $p : X longrightarrow X/sim_f$ be the quotient map where $X/sim_f$ is endowed with quotient topology. Then it is not hard to see that $tilde f circ p = f$. SInce $f$ is continuous so by Universal Mapping Property we can say that $tilde f$ is continuous.
Now my question is can we say that $tilde f$ is a homeomorphism? My instructor said that it is true and we were given it to prove.
What I have tried is as follows $:$
In order to show that $tilde f$ is a homeomorphism we need only to prove that it is open. So I take an open subset $tilde V subset X/sim_f$. We need to prove that $tilde f (tilde V)$ is open in $Y$. Now what is $tilde f (tilde V)$? I have found that $tilde f (tilde V) = f(S)$. Where $$S = cup_A in tilde V A.$$ Now since $tilde V$ is open in $X/sim_f$ so $p^-1 ( tilde V )$ is open in $X$. Now what is $p^-1 (tildeV)$? I have found that $p^-1 (tildeV) = S$. So we have to prove that $f(S)$ is an open subset of $Y$, where $S$ is an open subset of $X$.
I tried to prove it by contradiction. If not let $f(S)$ is not an open subset of $Y$. But that implies $f^-1(f(S))$ is not an open subset of $X$. If $f$ is one-to-one we are clearly done since then $f^-1(f(S))=S$ which leads us to a contradiction because $S$ was an open subset of $X$. But if $f$ is not one-to-one I don't know how to proceed. Please help me in this regard.
Thank you very much.
general-topology quotient-spaces
asked 13 hours ago
Debabrata Chattopadhyay.
8311
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X$ and $Y$ be two topological spaces. Suppose $f:X longrightarrow Y$ be a quotient map. Consider the equivalence class of $X$ formed by the disjoint non-empty fibres of $f$. Let $X/sim_f$ be the set of all these non-empty disjoint classes of $X$. Consider the map $tilde f : X/sim_f longrightarrow Y$ defined by $tilde f ([x]) = f(x),$ $x in X$. Then this map is well-defined and one-to-one by the construction and also since $f$ is onto, $tilde f$ is onto. Let $p : X longrightarrow X/sim_f$ be the quotient map where $X/sim_f$ is endowed with quotient topology. Then it is not hard to see that $tilde f circ p = f$. SInce $f$ is continuous so by Universal Mapping Property we can say that $tilde f$ is continuous.
Now my question is can we say that $tilde f$ is a homeomorphism? My instructor said that it is true and we were given it to prove.
What I have tried is as follows $:$
In order to show that $tilde f$ is a homeomorphism we need only to prove that it is open. So I take an open subset $tilde V subset X/sim_f$. We need to prove that $tilde f (tilde V)$ is open in $Y$. Now what is $tilde f (tilde V)$? I have found that $tilde f (tilde V) = f(S)$. Where $$S = cup_A in tilde V A.$$ Now since $tilde V$ is open in $X/sim_f$ so $p^-1 ( tilde V )$ is open in $X$. Now what is $p^-1 (tildeV)$? I have found that $p^-1 (tildeV) = S$. So we have to prove that $f(S)$ is an open subset of $Y$, where $S$ is an open subset of $X$.
I tried to prove it by contradiction. If not let $f(S)$ is not an open subset of $Y$. But that implies $f^-1(f(S))$ is not an open subset of $X$. If $f$ is one-to-one we are clearly done since then $f^-1(f(S))=S$ which leads us to a contradiction because $S$ was an open subset of $X$. But if $f$ is not one-to-one I don't know how to proceed. Please help me in this regard.
Thank you very much.
general-topology quotient-spaces
asked 13 hours ago
Debabrata Chattopadhyay.
8311
Let $X$ and $Y$ be two topological spaces. Suppose $f:X longrightarrow Y$ be a quotient map. Consider the equivalence class of $X$ formed by the disjoint non-empty fibres of $f$. Let $X/sim_f$ be the set of all these non-empty disjoint classes of $X$. Consider the map $tilde f : X/sim_f longrightarrow Y$ defined by $tilde f ([x]) = f(x),$ $x in X$. Then this map is well-defined and one-to-one by the construction and also since $f$ is onto, $tilde f$ is onto. Let $p : X longrightarrow X/sim_f$ be the quotient map where $X/sim_f$ is endowed with quotient topology. Then it is not hard to see that $tilde f circ p = f$. SInce $f$ is continuous so by Universal Mapping Property we can say that $tilde f$ is continuous.
Now my question is can we say that $tilde f$ is a homeomorphism? My instructor said that it is true and we were given it to prove.
What I have tried is as follows $:$
In order to show that $tilde f$ is a homeomorphism we need only to prove that it is open. So I take an open subset $tilde V subset X/sim_f$. We need to prove that $tilde f (tilde V)$ is open in $Y$. Now what is $tilde f (tilde V)$? I have found that $tilde f (tilde V) = f(S)$. Where $$S = cup_A in tilde V A.$$ Now since $tilde V$ is open in $X/sim_f$ so $p^-1 ( tilde V )$ is open in $X$. Now what is $p^-1 (tildeV)$? I have found that $p^-1 (tildeV) = S$. So we have to prove that $f(S)$ is an open subset of $Y$, where $S$ is an open subset of $X$.
I tried to prove it by contradiction. If not let $f(S)$ is not an open subset of $Y$. But that implies $f^-1(f(S))$ is not an open subset of $X$. If $f$ is one-to-one we are clearly done since then $f^-1(f(S))=S$ which leads us to a contradiction because $S$ was an open subset of $X$. But if $f$ is not one-to-one I don't know how to proceed. Please help me in this regard.
Thank you very much.
general-topology quotient-spaces
asked 13 hours ago
Debabrata Chattopadhyay.
8311
edited 13 hours ago
asked 13 hours ago
Debabrata Chattopadhyay.
8311
asked 13 hours ago
Debabrata Chattopadhyay.
8311
asked 13 hours ago
Debabrata Chattopadhyay.
8311
8311
add a comment |Â
add a comment |Â
1 Answer
1
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up vote
1
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accepted
You prove that $tildef$ is open and that is o.k. However, it is unnecessary. You have two quotient maps $f : X to Y$, $p : X to X / sim f$ and a bijection $tildef : X / sim f to Y$ such that $tildef circ p = f$. You correctly conclude that $tildef$ is continuous by the universal property of the quotient map $p$. The same argument shows that $tildef^-1$ is continuous because $tildef^-1 circ f = p$.
answered 12 hours ago
Paul Frost
3,308320
Exactly. I have overlooked it. Many many thanks @Paul Frost for helping me by pointing this out.
– Debabrata Chattopadhyay.
12 hours ago
So it follows from what I proved that if $f$ is open,continuous and onto then $tilde f$ is a homeomorphism. Isn't it so @Paul Frost?
– Debabrata Chattopadhyay.
12 hours ago
Oh! What a result I have proved on my own. I can't believe it. That is the joy of mathematics.
– Debabrata Chattopadhyay.
12 hours ago
1
Yes, by construction $tildef$ is a continuous bijection and it remains to show that it is open. You didn't finish the proof, but the approach is absolutely correct. Here is an easy proof: Let $tildeV$ be open in $X / sim f$. Then $f^-1(tildef(tildeV)) = (tildef circ p)^-1(tildef(tildeV)) = p^-1(tildef^-1(tildef(tildeV))) = p^-1(tildeV)$ which is open in $X$. Therefore $tildef(tildeV)$ is open in $Y$.
– Paul Frost
12 hours ago
1
Yes, that is correct!
– Paul Frost
12 hours ago
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You prove that $tildef$ is open and that is o.k. However, it is unnecessary. You have two quotient maps $f : X to Y$, $p : X to X / sim f$ and a bijection $tildef : X / sim f to Y$ such that $tildef circ p = f$. You correctly conclude that $tildef$ is continuous by the universal property of the quotient map $p$. The same argument shows that $tildef^-1$ is continuous because $tildef^-1 circ f = p$.
answered 12 hours ago
Paul Frost
3,308320
Exactly. I have overlooked it. Many many thanks @Paul Frost for helping me by pointing this out.
– Debabrata Chattopadhyay.
12 hours ago
So it follows from what I proved that if $f$ is open,continuous and onto then $tilde f$ is a homeomorphism. Isn't it so @Paul Frost?
– Debabrata Chattopadhyay.
12 hours ago
Oh! What a result I have proved on my own. I can't believe it. That is the joy of mathematics.
– Debabrata Chattopadhyay.
12 hours ago
1
Yes, by construction $tildef$ is a continuous bijection and it remains to show that it is open. You didn't finish the proof, but the approach is absolutely correct. Here is an easy proof: Let $tildeV$ be open in $X / sim f$. Then $f^-1(tildef(tildeV)) = (tildef circ p)^-1(tildef(tildeV)) = p^-1(tildef^-1(tildef(tildeV))) = p^-1(tildeV)$ which is open in $X$. Therefore $tildef(tildeV)$ is open in $Y$.
– Paul Frost
12 hours ago
1
Yes, that is correct!
– Paul Frost
12 hours ago
 |Â
show 5 more comments
up vote
1
down vote
accepted
You prove that $tildef$ is open and that is o.k. However, it is unnecessary. You have two quotient maps $f : X to Y$, $p : X to X / sim f$ and a bijection $tildef : X / sim f to Y$ such that $tildef circ p = f$. You correctly conclude that $tildef$ is continuous by the universal property of the quotient map $p$. The same argument shows that $tildef^-1$ is continuous because $tildef^-1 circ f = p$.
answered 12 hours ago
Paul Frost
3,308320
Exactly. I have overlooked it. Many many thanks @Paul Frost for helping me by pointing this out.
– Debabrata Chattopadhyay.
12 hours ago
So it follows from what I proved that if $f$ is open,continuous and onto then $tilde f$ is a homeomorphism. Isn't it so @Paul Frost?
– Debabrata Chattopadhyay.
12 hours ago
Oh! What a result I have proved on my own. I can't believe it. That is the joy of mathematics.
– Debabrata Chattopadhyay.
12 hours ago
1
Yes, by construction $tildef$ is a continuous bijection and it remains to show that it is open. You didn't finish the proof, but the approach is absolutely correct. Here is an easy proof: Let $tildeV$ be open in $X / sim f$. Then $f^-1(tildef(tildeV)) = (tildef circ p)^-1(tildef(tildeV)) = p^-1(tildef^-1(tildef(tildeV))) = p^-1(tildeV)$ which is open in $X$. Therefore $tildef(tildeV)$ is open in $Y$.
– Paul Frost
12 hours ago
1
Yes, that is correct!
– Paul Frost
12 hours ago
 |Â
show 5 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You prove that $tildef$ is open and that is o.k. However, it is unnecessary. You have two quotient maps $f : X to Y$, $p : X to X / sim f$ and a bijection $tildef : X / sim f to Y$ such that $tildef circ p = f$. You correctly conclude that $tildef$ is continuous by the universal property of the quotient map $p$. The same argument shows that $tildef^-1$ is continuous because $tildef^-1 circ f = p$.
answered 12 hours ago
Paul Frost
3,308320
You prove that $tildef$ is open and that is o.k. However, it is unnecessary. You have two quotient maps $f : X to Y$, $p : X to X / sim f$ and a bijection $tildef : X / sim f to Y$ such that $tildef circ p = f$. You correctly conclude that $tildef$ is continuous by the universal property of the quotient map $p$. The same argument shows that $tildef^-1$ is continuous because $tildef^-1 circ f = p$.
answered 12 hours ago
Paul Frost
3,308320
answered 12 hours ago
Paul Frost
3,308320
answered 12 hours ago
Paul Frost
3,308320
answered 12 hours ago
Paul Frost
3,308320
3,308320
Exactly. I have overlooked it. Many many thanks @Paul Frost for helping me by pointing this out.
– Debabrata Chattopadhyay.
12 hours ago
So it follows from what I proved that if $f$ is open,continuous and onto then $tilde f$ is a homeomorphism. Isn't it so @Paul Frost?
– Debabrata Chattopadhyay.
12 hours ago
Oh! What a result I have proved on my own. I can't believe it. That is the joy of mathematics.
– Debabrata Chattopadhyay.
12 hours ago
1
Yes, by construction $tildef$ is a continuous bijection and it remains to show that it is open. You didn't finish the proof, but the approach is absolutely correct. Here is an easy proof: Let $tildeV$ be open in $X / sim f$. Then $f^-1(tildef(tildeV)) = (tildef circ p)^-1(tildef(tildeV)) = p^-1(tildef^-1(tildef(tildeV))) = p^-1(tildeV)$ which is open in $X$. Therefore $tildef(tildeV)$ is open in $Y$.
– Paul Frost
12 hours ago
1
Yes, that is correct!
– Paul Frost
12 hours ago
 |Â
show 5 more comments
Exactly. I have overlooked it. Many many thanks @Paul Frost for helping me by pointing this out.
– Debabrata Chattopadhyay.
12 hours ago
So it follows from what I proved that if $f$ is open,continuous and onto then $tilde f$ is a homeomorphism. Isn't it so @Paul Frost?
– Debabrata Chattopadhyay.
12 hours ago
Oh! What a result I have proved on my own. I can't believe it. That is the joy of mathematics.
– Debabrata Chattopadhyay.
12 hours ago
1
Yes, by construction $tildef$ is a continuous bijection and it remains to show that it is open. You didn't finish the proof, but the approach is absolutely correct. Here is an easy proof: Let $tildeV$ be open in $X / sim f$. Then $f^-1(tildef(tildeV)) = (tildef circ p)^-1(tildef(tildeV)) = p^-1(tildef^-1(tildef(tildeV))) = p^-1(tildeV)$ which is open in $X$. Therefore $tildef(tildeV)$ is open in $Y$.
– Paul Frost
12 hours ago
1
Yes, that is correct!
– Paul Frost
12 hours ago
Exactly. I have overlooked it. Many many thanks @Paul Frost for helping me by pointing this out.
– Debabrata Chattopadhyay.
12 hours ago
Exactly. I have overlooked it. Many many thanks @Paul Frost for helping me by pointing this out.
– Debabrata Chattopadhyay.
12 hours ago
So it follows from what I proved that if $f$ is open,continuous and onto then $tilde f$ is a homeomorphism. Isn't it so @Paul Frost?
– Debabrata Chattopadhyay.
12 hours ago
So it follows from what I proved that if $f$ is open,continuous and onto then $tilde f$ is a homeomorphism. Isn't it so @Paul Frost?
– Debabrata Chattopadhyay.
12 hours ago
Oh! What a result I have proved on my own. I can't believe it. That is the joy of mathematics.
– Debabrata Chattopadhyay.
12 hours ago
Oh! What a result I have proved on my own. I can't believe it. That is the joy of mathematics.
– Debabrata Chattopadhyay.
12 hours ago
1
1
Yes, by construction $tildef$ is a continuous bijection and it remains to show that it is open. You didn't finish the proof, but the approach is absolutely correct. Here is an easy proof: Let $tildeV$ be open in $X / sim f$. Then $f^-1(tildef(tildeV)) = (tildef circ p)^-1(tildef(tildeV)) = p^-1(tildef^-1(tildef(tildeV))) = p^-1(tildeV)$ which is open in $X$. Therefore $tildef(tildeV)$ is open in $Y$.
– Paul Frost
12 hours ago
Yes, by construction $tildef$ is a continuous bijection and it remains to show that it is open. You didn't finish the proof, but the approach is absolutely correct. Here is an easy proof: Let $tildeV$ be open in $X / sim f$. Then $f^-1(tildef(tildeV)) = (tildef circ p)^-1(tildef(tildeV)) = p^-1(tildef^-1(tildef(tildeV))) = p^-1(tildeV)$ which is open in $X$. Therefore $tildef(tildeV)$ is open in $Y$.
– Paul Frost
12 hours ago
1
1
Yes, that is correct!
– Paul Frost
12 hours ago
Yes, that is correct!
– Paul Frost
12 hours ago
 |Â
show 5 more comments
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