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I was looking for this, but I can't find anything.

Problem. Let $A, B$ two compact surfaces such that $A\#B \cong S^2$ then $A\cong B \cong S^2$.

I considerd infinite connected sum $A\#B\#A\dotsc$ and $B\#A\#B\dotsc$ These are homeomorphic to $\mathbb{R}^2$ and with $\{\infty\}$ to $S^2$, but I cant finalize the prove.

Can you get me a hint? Thank

Edit. Can I prove it using this? Connected sum of non orientable surfaces is non orienable, then A and B are orientables, in fact, are respectively connected sum of n and m torus. Hence we also know that connected sum of A and B is homeomorphic to the connected sum of $n+m$ torus. Therefore $m+n=0$ and $m,n\geq 0$ so $m=n=0$.

Adam Hughes
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Rafael Gonzalez Lopez
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2 Answers2

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Compact surfaces form a (commutative) monoid under connected sum where $S^2$ is the identity. If you don't already know that, just consider what connected sum does, it cuts out a disk and identifies the boundary, so if you have anything non-orientable it stays that way, and if you have any handles, it only increases the number of handles.

Adam Hughes
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As you have mentioned, $A\#(B\#A\#B\#\cdots)\simeq \mathbb{R}^2$ and $B\#A\#B\#\cdots\simeq \mathbb{R}^2$.

But observe that if $X$ is a compact surface, $X\#\mathbb{R}^2 \simeq X\setminus\{x_0\}$ (=$X$ minus a point) Therefore, it means that $A$ minus a point is homeomorphic to $\mathbb{R}^2$, and then it is easy to conclude that $A \simeq S^2$. That $B\simeq S^2$ follows easily by symmetry.

Henry
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  • That was the point! Thank you for comment, but I would rather prove it without using inifinite sum ^^ – Rafael Gonzalez Lopez May 03 '17 at 12:27
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    @RafaelGonzalezLopez In fact, it is a very famous argument due to Barry Mazur. Exactly the same argument can be applied to prove some other theorems, such as "the only invertible elements in the monoid of knots and their knot sums is the unknot". See, for instance, http://www.math.stonybrook.edu/~jack/MilnorOslo-print.pdf or https://en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur_swindle – Henry May 03 '17 at 12:30