If $X$ is compact and Hausdorff, and $f:Xrightarrow Y$ is continuous, closed, and surjective, prove $Y$ is...












0












$begingroup$



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







share|cite|improve this question











$endgroup$

















    0












    $begingroup$



    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







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      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







      share|cite|improve this question











      $endgroup$





      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 13 at 0:49







      Arbutus

















      asked Jan 13 at 0:43









      ArbutusArbutus

      755715




      755715






















          3 Answers
          3






          active

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          0












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah, of course. Thank you!
            $endgroup$
            – Arbutus
            Jan 13 at 0:55



















          1












          $begingroup$

          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).






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            $f$ is not injective, so you can have $y_1in f(X/U_1)$.






            share|cite|improve this answer









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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              0












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Ah, of course. Thank you!
                $endgroup$
                – Arbutus
                Jan 13 at 0:55
















              0












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Ah, of course. Thank you!
                $endgroup$
                – Arbutus
                Jan 13 at 0:55














              0












              0








              0





              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$



              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.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              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


















              • $begingroup$
                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











              1












              $begingroup$

              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).






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                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).






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  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).






                  share|cite|improve this answer











                  $endgroup$



                  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).







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 13 at 15:36

























                  answered Jan 13 at 14:07









                  Henno BrandsmaHenno Brandsma

                  116k349127




                  116k349127























                      0












                      $begingroup$

                      $f$ is not injective, so you can have $y_1in f(X/U_1)$.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        $f$ is not injective, so you can have $y_1in f(X/U_1)$.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          $f$ is not injective, so you can have $y_1in f(X/U_1)$.






                          share|cite|improve this answer









                          $endgroup$



                          $f$ is not injective, so you can have $y_1in f(X/U_1)$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 13 at 0:51









                          Tsemo AristideTsemo Aristide

                          60.7k11446




                          60.7k11446






























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