Connected sum of two non homeomorphic surfaces

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I must solve this exercise:




$S_1, S_2, S_3, S_4$ are four pairwise non-homeomorphic surfaces which
are connected and compact.
Is it correct that $S_1#S_2$ is not homeomorphic to $S_3#S_4$ ?




I tried to find a counterexample in order to prove that it is not correct.
My approach is the following:



  • I choose a surface $S_1$

  • Classification theorem says that $S_1$ is homeomorphic to a sphere or to a connected sum of tori or to a connected sum of projective planes.

  • I decide $S_1$ to be homeomorphic to a sum of three projective planes: $S_1 cong mathbbP # mathbbP #mathbbP$

  • I take $S_2$ = sphere, $S_3$ = torus T, $S_4$ = projective plane $mathbbP$

  • I have four pairwise non homeomorphic functions, but $S_1 # S_2 =S_1$ and $T#mathbbP cong mathbbP #mathbbP #mathbbP$ so $S_1 # S_2 cong T#mathbbP$

In this way I found a counterexamle which proved that there exists a case in which $S_1, S_2, S_3, S_4$ are four pairwise non-homeomorphic surfaces but $S_1 # S_2 cong T#mathbbP$ ($A implies neg B$). So the question in the exercise is false.



Is my answer right?






general-topology surfaces projective-space







share|cite|improve this question























  • But you make a specific choice for you surfaces. So this proof isn't general at all.
    – NDewolf
    Aug 30 at 8:58










  • I thought about this as a counterexample which should prove that $neg B implies A rightarrowleftarrow $
    – Phi_24
    Aug 30 at 9:04











  • I don't see how this is a counterexample. You have proven that $S_1#S_2$ is homeomorphic to $S_3#S_4$. Or am i misunderstanding your initial statement?
    – NDewolf
    Aug 30 at 9:15











  • I think I found a case in which $S_1, S_2, S_3, S_4$ are non homeomorphic but $S_1 # S_2$ is homeomorphic to $S_3 # S4$, so "$S_1, S_2, S_3, S_4$ non homeomorphic" $nRightarrow S_1 # S_2 cong S_3 # S4$
    – Phi_24
    Aug 30 at 9:21











  • But you found a case in which the implication seems to be valid. To use contraposition, as you wanted, you should find a case in which $S_1#S_2 notcong S_3#S_4implies S_1, S_2, S_3, S_4$ not mutually non-homeomorphic.
    – NDewolf
    Aug 30 at 9:25















up vote
0
down vote

favorite












I must solve this exercise:




$S_1, S_2, S_3, S_4$ are four pairwise non-homeomorphic surfaces which
are connected and compact.
Is it correct that $S_1#S_2$ is not homeomorphic to $S_3#S_4$ ?




I tried to find a counterexample in order to prove that it is not correct.
My approach is the following:



  • I choose a surface $S_1$

  • Classification theorem says that $S_1$ is homeomorphic to a sphere or to a connected sum of tori or to a connected sum of projective planes.

  • I decide $S_1$ to be homeomorphic to a sum of three projective planes: $S_1 cong mathbbP # mathbbP #mathbbP$

  • I take $S_2$ = sphere, $S_3$ = torus T, $S_4$ = projective plane $mathbbP$

  • I have four pairwise non homeomorphic functions, but $S_1 # S_2 =S_1$ and $T#mathbbP cong mathbbP #mathbbP #mathbbP$ so $S_1 # S_2 cong T#mathbbP$

In this way I found a counterexamle which proved that there exists a case in which $S_1, S_2, S_3, S_4$ are four pairwise non-homeomorphic surfaces but $S_1 # S_2 cong T#mathbbP$ ($A implies neg B$). So the question in the exercise is false.



Is my answer right?






general-topology surfaces projective-space







share|cite|improve this question























  • But you make a specific choice for you surfaces. So this proof isn't general at all.
    – NDewolf
    Aug 30 at 8:58










  • I thought about this as a counterexample which should prove that $neg B implies A rightarrowleftarrow $
    – Phi_24
    Aug 30 at 9:04











  • I don't see how this is a counterexample. You have proven that $S_1#S_2$ is homeomorphic to $S_3#S_4$. Or am i misunderstanding your initial statement?
    – NDewolf
    Aug 30 at 9:15











  • I think I found a case in which $S_1, S_2, S_3, S_4$ are non homeomorphic but $S_1 # S_2$ is homeomorphic to $S_3 # S4$, so "$S_1, S_2, S_3, S_4$ non homeomorphic" $nRightarrow S_1 # S_2 cong S_3 # S4$
    – Phi_24
    Aug 30 at 9:21











  • But you found a case in which the implication seems to be valid. To use contraposition, as you wanted, you should find a case in which $S_1#S_2 notcong S_3#S_4implies S_1, S_2, S_3, S_4$ not mutually non-homeomorphic.
    – NDewolf
    Aug 30 at 9:25













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I must solve this exercise:




$S_1, S_2, S_3, S_4$ are four pairwise non-homeomorphic surfaces which
are connected and compact.
Is it correct that $S_1#S_2$ is not homeomorphic to $S_3#S_4$ ?




I tried to find a counterexample in order to prove that it is not correct.
My approach is the following:



  • I choose a surface $S_1$

  • Classification theorem says that $S_1$ is homeomorphic to a sphere or to a connected sum of tori or to a connected sum of projective planes.

  • I decide $S_1$ to be homeomorphic to a sum of three projective planes: $S_1 cong mathbbP # mathbbP #mathbbP$

  • I take $S_2$ = sphere, $S_3$ = torus T, $S_4$ = projective plane $mathbbP$

  • I have four pairwise non homeomorphic functions, but $S_1 # S_2 =S_1$ and $T#mathbbP cong mathbbP #mathbbP #mathbbP$ so $S_1 # S_2 cong T#mathbbP$

In this way I found a counterexamle which proved that there exists a case in which $S_1, S_2, S_3, S_4$ are four pairwise non-homeomorphic surfaces but $S_1 # S_2 cong T#mathbbP$ ($A implies neg B$). So the question in the exercise is false.



Is my answer right?






general-topology surfaces projective-space







share|cite|improve this question















I must solve this exercise:




$S_1, S_2, S_3, S_4$ are four pairwise non-homeomorphic surfaces which
are connected and compact.
Is it correct that $S_1#S_2$ is not homeomorphic to $S_3#S_4$ ?




I tried to find a counterexample in order to prove that it is not correct.
My approach is the following:



  • I choose a surface $S_1$

  • Classification theorem says that $S_1$ is homeomorphic to a sphere or to a connected sum of tori or to a connected sum of projective planes.

  • I decide $S_1$ to be homeomorphic to a sum of three projective planes: $S_1 cong mathbbP # mathbbP #mathbbP$

  • I take $S_2$ = sphere, $S_3$ = torus T, $S_4$ = projective plane $mathbbP$

  • I have four pairwise non homeomorphic functions, but $S_1 # S_2 =S_1$ and $T#mathbbP cong mathbbP #mathbbP #mathbbP$ so $S_1 # S_2 cong T#mathbbP$

In this way I found a counterexamle which proved that there exists a case in which $S_1, S_2, S_3, S_4$ are four pairwise non-homeomorphic surfaces but $S_1 # S_2 cong T#mathbbP$ ($A implies neg B$). So the question in the exercise is false.



Is my answer right?






general-topology surfaces projective-space




general-topology surfaces projective-space






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 30 at 9:46

























asked Aug 30 at 8:46









Phi_24

675




675











  • But you make a specific choice for you surfaces. So this proof isn't general at all.
    – NDewolf
    Aug 30 at 8:58










  • I thought about this as a counterexample which should prove that $neg B implies A rightarrowleftarrow $
    – Phi_24
    Aug 30 at 9:04











  • I don't see how this is a counterexample. You have proven that $S_1#S_2$ is homeomorphic to $S_3#S_4$. Or am i misunderstanding your initial statement?
    – NDewolf
    Aug 30 at 9:15











  • I think I found a case in which $S_1, S_2, S_3, S_4$ are non homeomorphic but $S_1 # S_2$ is homeomorphic to $S_3 # S4$, so "$S_1, S_2, S_3, S_4$ non homeomorphic" $nRightarrow S_1 # S_2 cong S_3 # S4$
    – Phi_24
    Aug 30 at 9:21











  • But you found a case in which the implication seems to be valid. To use contraposition, as you wanted, you should find a case in which $S_1#S_2 notcong S_3#S_4implies S_1, S_2, S_3, S_4$ not mutually non-homeomorphic.
    – NDewolf
    Aug 30 at 9:25

















  • But you make a specific choice for you surfaces. So this proof isn't general at all.
    – NDewolf
    Aug 30 at 8:58










  • I thought about this as a counterexample which should prove that $neg B implies A rightarrowleftarrow $
    – Phi_24
    Aug 30 at 9:04











  • I don't see how this is a counterexample. You have proven that $S_1#S_2$ is homeomorphic to $S_3#S_4$. Or am i misunderstanding your initial statement?
    – NDewolf
    Aug 30 at 9:15











  • I think I found a case in which $S_1, S_2, S_3, S_4$ are non homeomorphic but $S_1 # S_2$ is homeomorphic to $S_3 # S4$, so "$S_1, S_2, S_3, S_4$ non homeomorphic" $nRightarrow S_1 # S_2 cong S_3 # S4$
    – Phi_24
    Aug 30 at 9:21











  • But you found a case in which the implication seems to be valid. To use contraposition, as you wanted, you should find a case in which $S_1#S_2 notcong S_3#S_4implies S_1, S_2, S_3, S_4$ not mutually non-homeomorphic.
    – NDewolf
    Aug 30 at 9:25
















But you make a specific choice for you surfaces. So this proof isn't general at all.
– NDewolf
Aug 30 at 8:58




But you make a specific choice for you surfaces. So this proof isn't general at all.
– NDewolf
Aug 30 at 8:58












I thought about this as a counterexample which should prove that $neg B implies A rightarrowleftarrow $
– Phi_24
Aug 30 at 9:04





I thought about this as a counterexample which should prove that $neg B implies A rightarrowleftarrow $
– Phi_24
Aug 30 at 9:04













I don't see how this is a counterexample. You have proven that $S_1#S_2$ is homeomorphic to $S_3#S_4$. Or am i misunderstanding your initial statement?
– NDewolf
Aug 30 at 9:15





I don't see how this is a counterexample. You have proven that $S_1#S_2$ is homeomorphic to $S_3#S_4$. Or am i misunderstanding your initial statement?
– NDewolf
Aug 30 at 9:15













I think I found a case in which $S_1, S_2, S_3, S_4$ are non homeomorphic but $S_1 # S_2$ is homeomorphic to $S_3 # S4$, so "$S_1, S_2, S_3, S_4$ non homeomorphic" $nRightarrow S_1 # S_2 cong S_3 # S4$
– Phi_24
Aug 30 at 9:21





I think I found a case in which $S_1, S_2, S_3, S_4$ are non homeomorphic but $S_1 # S_2$ is homeomorphic to $S_3 # S4$, so "$S_1, S_2, S_3, S_4$ non homeomorphic" $nRightarrow S_1 # S_2 cong S_3 # S4$
– Phi_24
Aug 30 at 9:21













But you found a case in which the implication seems to be valid. To use contraposition, as you wanted, you should find a case in which $S_1#S_2 notcong S_3#S_4implies S_1, S_2, S_3, S_4$ not mutually non-homeomorphic.
– NDewolf
Aug 30 at 9:25





But you found a case in which the implication seems to be valid. To use contraposition, as you wanted, you should find a case in which $S_1#S_2 notcong S_3#S_4implies S_1, S_2, S_3, S_4$ not mutually non-homeomorphic.
– NDewolf
Aug 30 at 9:25
















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