If $X$ is compact and Hausdorff, and $f:Xrightarrow Y$ is continuous, closed, and surjective, prove $Y$ is...
$begingroup$
Let $X$ be a compact Hausdorff space. If $f:Xrightarrow Y$ is continuous, closed, and surjective, prove that $Y$ is Hausdorff.
I'm wondering if/where the compactness of $X$ is needed. I have the following 'proof', but I'm rather skeptical as it
doesn't depend on the compactness of $X$. Where is the mistake in my proof?
Proof: Let $y_1,y_2in Y$ be distinct. Then there are $x_1,x_2in X$ such that $f(x_i)=y_i$. Since $X$ is Hausdorff, there exist disjoint open neighborhoods $U_1, U_2$ of $x_1$ and $x_2$, respectively. As $f$ is closed, $f(Xbackslash U_i)subseteq Y$ are also closed. Thus, if $V_i:=Ybackslash f(Xbackslash U_i)$, then $V_1$ and $V_2$ are disjoint open neighborhoods of $y_1$ and $y_2$, respectively, where disjointness follows from
begin{align*}
Ybackslash( V_1cap V_2)&=Ybackslash V_1cup Ybackslash V_2 \
&=f(Xbackslash U_1)cup f(Xbackslash U_2)\
&=f(Xbackslash U_1cup Xbackslash U_2)\
&=f(Xbackslash (U_1cap U_2))\
&=f(X)\
&=Y.
end{align*}
general-topology proof-verification continuity compactness
asked Jan 13 at 0:43
ArbutusArbutus
755715
$endgroup$
add a comment |
$begingroup$
Let $X$ be a compact Hausdorff space. If $f:Xrightarrow Y$ is continuous, closed, and surjective, prove that $Y$ is Hausdorff.
I'm wondering if/where the compactness of $X$ is needed. I have the following 'proof', but I'm rather skeptical as it
doesn't depend on the compactness of $X$. Where is the mistake in my proof?
Proof: Let $y_1,y_2in Y$ be distinct. Then there are $x_1,x_2in X$ such that $f(x_i)=y_i$. Since $X$ is Hausdorff, there exist disjoint open neighborhoods $U_1, U_2$ of $x_1$ and $x_2$, respectively. As $f$ is closed, $f(Xbackslash U_i)subseteq Y$ are also closed. Thus, if $V_i:=Ybackslash f(Xbackslash U_i)$, then $V_1$ and $V_2$ are disjoint open neighborhoods of $y_1$ and $y_2$, respectively, where disjointness follows from
begin{align*}
Ybackslash( V_1cap V_2)&=Ybackslash V_1cup Ybackslash V_2 \
&=f(Xbackslash U_1)cup f(Xbackslash U_2)\
&=f(Xbackslash U_1cup Xbackslash U_2)\
&=f(Xbackslash (U_1cap U_2))\
&=f(X)\
&=Y.
end{align*}
general-topology proof-verification continuity compactness
asked Jan 13 at 0:43
ArbutusArbutus
755715
$endgroup$
add a comment |
$begingroup$
Let $X$ be a compact Hausdorff space. If $f:Xrightarrow Y$ is continuous, closed, and surjective, prove that $Y$ is Hausdorff.
I'm wondering if/where the compactness of $X$ is needed. I have the following 'proof', but I'm rather skeptical as it
doesn't depend on the compactness of $X$. Where is the mistake in my proof?
Proof: Let $y_1,y_2in Y$ be distinct. Then there are $x_1,x_2in X$ such that $f(x_i)=y_i$. Since $X$ is Hausdorff, there exist disjoint open neighborhoods $U_1, U_2$ of $x_1$ and $x_2$, respectively. As $f$ is closed, $f(Xbackslash U_i)subseteq Y$ are also closed. Thus, if $V_i:=Ybackslash f(Xbackslash U_i)$, then $V_1$ and $V_2$ are disjoint open neighborhoods of $y_1$ and $y_2$, respectively, where disjointness follows from
begin{align*}
Ybackslash( V_1cap V_2)&=Ybackslash V_1cup Ybackslash V_2 \
&=f(Xbackslash U_1)cup f(Xbackslash U_2)\
&=f(Xbackslash U_1cup Xbackslash U_2)\
&=f(Xbackslash (U_1cap U_2))\
&=f(X)\
&=Y.
end{align*}
general-topology proof-verification continuity compactness
asked Jan 13 at 0:43
ArbutusArbutus
755715
$endgroup$
Let $X$ be a compact Hausdorff space. If $f:Xrightarrow Y$ is continuous, closed, and surjective, prove that $Y$ is Hausdorff.
I'm wondering if/where the compactness of $X$ is needed. I have the following 'proof', but I'm rather skeptical as it
doesn't depend on the compactness of $X$. Where is the mistake in my proof?
Proof: Let $y_1,y_2in Y$ be distinct. Then there are $x_1,x_2in X$ such that $f(x_i)=y_i$. Since $X$ is Hausdorff, there exist disjoint open neighborhoods $U_1, U_2$ of $x_1$ and $x_2$, respectively. As $f$ is closed, $f(Xbackslash U_i)subseteq Y$ are also closed. Thus, if $V_i:=Ybackslash f(Xbackslash U_i)$, then $V_1$ and $V_2$ are disjoint open neighborhoods of $y_1$ and $y_2$, respectively, where disjointness follows from
begin{align*}
Ybackslash( V_1cap V_2)&=Ybackslash V_1cup Ybackslash V_2 \
&=f(Xbackslash U_1)cup f(Xbackslash U_2)\
&=f(Xbackslash U_1cup Xbackslash U_2)\
&=f(Xbackslash (U_1cap U_2))\
&=f(X)\
&=Y.
end{align*}
general-topology proof-verification continuity compactness
general-topology proof-verification continuity compactness
asked Jan 13 at 0:43
ArbutusArbutus
755715
asked Jan 13 at 0:43
ArbutusArbutus
755715
edited Jan 13 at 0:49
Arbutus
asked Jan 13 at 0:43
ArbutusArbutus
755715
asked Jan 13 at 0:43
ArbutusArbutus
755715
asked Jan 13 at 0:43
ArbutusArbutus
755715
755715
add a comment |
add a comment |
3 Answers
3
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The mistake is in claiming that $y_1 in V_1$ and $y_2 in V_2$. You cannot say this unless $f$ is injective.
answered Jan 13 at 0:50
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
Ah, of course. Thank you!
$endgroup$
– Arbutus
Jan 13 at 0:55
add a comment |
$begingroup$
The non-injectivity failure in your current proof has already been pointed out. Now instead of disjoint neighbourhoods $U_1,U_2$ of $x_1$ and $x_2$, use disjoint neighbourhoods of the closed sets $F_i = f^{-1}[{y_i}]$ ($i=1,2$) of $X$ and the same construction of open sets. This does use compactness (why?) and we do get neighbourhoods of $y_i$ ( this needs a small argument).
A more general theorem:
Suppose that $f: X to Y$ is surjective, continuous, closed and has compact fibres (all $f^{-1}[{y}], y in Y$ are compact (such a map is called perfect)) then if $X$ is Hausdorff, then so is $Y$.
This captures the essence of the proof and does not require $X$ to be compact (we need the compactness of the fibres, really).
answered Jan 13 at 14:07
Henno BrandsmaHenno Brandsma
116k349127
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add a comment |
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$f$ is not injective, so you can have $y_1in f(X/U_1)$.
answered Jan 13 at 0:51
Tsemo AristideTsemo Aristide
60.7k11446
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
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$begingroup$
The mistake is in claiming that $y_1 in V_1$ and $y_2 in V_2$. You cannot say this unless $f$ is injective.
answered Jan 13 at 0:50
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
Ah, of course. Thank you!
$endgroup$
– Arbutus
Jan 13 at 0:55
add a comment |
$begingroup$
The mistake is in claiming that $y_1 in V_1$ and $y_2 in V_2$. You cannot say this unless $f$ is injective.
answered Jan 13 at 0:50
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
$begingroup$
Ah, of course. Thank you!
$endgroup$
– Arbutus
Jan 13 at 0:55
add a comment |
$begingroup$
The mistake is in claiming that $y_1 in V_1$ and $y_2 in V_2$. You cannot say this unless $f$ is injective.
answered Jan 13 at 0:50
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
$endgroup$
The mistake is in claiming that $y_1 in V_1$ and $y_2 in V_2$. You cannot say this unless $f$ is injective.
answered Jan 13 at 0:50
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Jan 13 at 0:50
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Jan 13 at 0:50
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
answered Jan 13 at 0:50
Kavi Rama MurthyKavi Rama Murthy
74.9k53270
74.9k53270
$begingroup$
Ah, of course. Thank you!
$endgroup$
– Arbutus
Jan 13 at 0:55
add a comment |
$begingroup$
Ah, of course. Thank you!
$endgroup$
– Arbutus
Jan 13 at 0:55
$begingroup$
Ah, of course. Thank you!
$endgroup$
– Arbutus
Jan 13 at 0:55
$begingroup$
Ah, of course. Thank you!
$endgroup$
– Arbutus
Jan 13 at 0:55
add a comment |
$begingroup$
The non-injectivity failure in your current proof has already been pointed out. Now instead of disjoint neighbourhoods $U_1,U_2$ of $x_1$ and $x_2$, use disjoint neighbourhoods of the closed sets $F_i = f^{-1}[{y_i}]$ ($i=1,2$) of $X$ and the same construction of open sets. This does use compactness (why?) and we do get neighbourhoods of $y_i$ ( this needs a small argument).
A more general theorem:
Suppose that $f: X to Y$ is surjective, continuous, closed and has compact fibres (all $f^{-1}[{y}], y in Y$ are compact (such a map is called perfect)) then if $X$ is Hausdorff, then so is $Y$.
This captures the essence of the proof and does not require $X$ to be compact (we need the compactness of the fibres, really).
answered Jan 13 at 14:07
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
The non-injectivity failure in your current proof has already been pointed out. Now instead of disjoint neighbourhoods $U_1,U_2$ of $x_1$ and $x_2$, use disjoint neighbourhoods of the closed sets $F_i = f^{-1}[{y_i}]$ ($i=1,2$) of $X$ and the same construction of open sets. This does use compactness (why?) and we do get neighbourhoods of $y_i$ ( this needs a small argument).
A more general theorem:
Suppose that $f: X to Y$ is surjective, continuous, closed and has compact fibres (all $f^{-1}[{y}], y in Y$ are compact (such a map is called perfect)) then if $X$ is Hausdorff, then so is $Y$.
This captures the essence of the proof and does not require $X$ to be compact (we need the compactness of the fibres, really).
answered Jan 13 at 14:07
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
The non-injectivity failure in your current proof has already been pointed out. Now instead of disjoint neighbourhoods $U_1,U_2$ of $x_1$ and $x_2$, use disjoint neighbourhoods of the closed sets $F_i = f^{-1}[{y_i}]$ ($i=1,2$) of $X$ and the same construction of open sets. This does use compactness (why?) and we do get neighbourhoods of $y_i$ ( this needs a small argument).
A more general theorem:
Suppose that $f: X to Y$ is surjective, continuous, closed and has compact fibres (all $f^{-1}[{y}], y in Y$ are compact (such a map is called perfect)) then if $X$ is Hausdorff, then so is $Y$.
This captures the essence of the proof and does not require $X$ to be compact (we need the compactness of the fibres, really).
answered Jan 13 at 14:07
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
The non-injectivity failure in your current proof has already been pointed out. Now instead of disjoint neighbourhoods $U_1,U_2$ of $x_1$ and $x_2$, use disjoint neighbourhoods of the closed sets $F_i = f^{-1}[{y_i}]$ ($i=1,2$) of $X$ and the same construction of open sets. This does use compactness (why?) and we do get neighbourhoods of $y_i$ ( this needs a small argument).
A more general theorem:
Suppose that $f: X to Y$ is surjective, continuous, closed and has compact fibres (all $f^{-1}[{y}], y in Y$ are compact (such a map is called perfect)) then if $X$ is Hausdorff, then so is $Y$.
This captures the essence of the proof and does not require $X$ to be compact (we need the compactness of the fibres, really).
answered Jan 13 at 14:07
Henno BrandsmaHenno Brandsma
116k349127
edited Jan 13 at 15:36
answered Jan 13 at 14:07
Henno BrandsmaHenno Brandsma
116k349127
answered Jan 13 at 14:07
Henno BrandsmaHenno Brandsma
116k349127
answered Jan 13 at 14:07
Henno BrandsmaHenno Brandsma
116k349127
116k349127
add a comment |
add a comment |
$begingroup$
$f$ is not injective, so you can have $y_1in f(X/U_1)$.
answered Jan 13 at 0:51
Tsemo AristideTsemo Aristide
60.7k11446
$endgroup$
add a comment |
$begingroup$
$f$ is not injective, so you can have $y_1in f(X/U_1)$.
answered Jan 13 at 0:51
Tsemo AristideTsemo Aristide
60.7k11446
$endgroup$
add a comment |
$begingroup$
$f$ is not injective, so you can have $y_1in f(X/U_1)$.
answered Jan 13 at 0:51
Tsemo AristideTsemo Aristide
60.7k11446
$endgroup$
$f$ is not injective, so you can have $y_1in f(X/U_1)$.
answered Jan 13 at 0:51
Tsemo AristideTsemo Aristide
60.7k11446
answered Jan 13 at 0:51
Tsemo AristideTsemo Aristide
60.7k11446
answered Jan 13 at 0:51
Tsemo AristideTsemo Aristide
60.7k11446
answered Jan 13 at 0:51
Tsemo AristideTsemo Aristide
60.7k11446
60.7k11446
add a comment |
add a comment |
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