Given a torus with an open disk removed and its antipodal boundary points identified (note that the boundary refers that of the removed disk, which is a $S^1$), what space is it?

At first I viewed the torus as a square with opposite edges identified, cut along a diagonal, and tried to glue the two parts together in a nice way, but it ended up more confusing. Then I tried to view the torus as a square glued with two strips (consider torus as two cylinders glued together over a square patch, [shown by the following picture from Basic Topology by Armstrong], I simply cut off the square and get three pieces), and the procedure amounts to replacing the two strips with two Mobius strips and identifying several points. With some struggle, I got the result: a real projective plane with two tangent disks replaced by two Mobius strips, which is not satisfactory.

I've been stuck since then. Please help. I'm not sure what kind of description can be considered good and clear. (Maybe the cell structure will do, but I couldn't determine it either.)

punctured torus


2 Answers 2


Claim: Let $X$ denote the space that you described. It's homeomorphic to $\Bbb{R}P^2\# T^2$, the connected sum of a real projective plane and a torus. (If I understand your question correctly)

I'll prove this by two ways:

Cut-And-Paste Argument:

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Or simply using algebraic notations about labelling scheme (of equivalence relation),

$$afab^{-1}fb^{-1}\mapsto b^{-1}afab^{-1}f \text{ by permutation}$$ $$\mapsto b^{-1}afx^{-1},\text{ }xab^{-1}f \text{ by cutting}$$ $$\mapsto x^{-1}b^{-1}af,\text{ }f^{-1}ba^{-1}x^{-1} \text{ by permuting the first part and reversing orientation of the second}$$ $$\mapsto x^{-1}b^{-1}aba^{-1}x^{-1}\text{ by pasting}$$ $$\mapsto b^{-1}aba^{-1}x^{-1}x^{-1}\text{ by permuting}$$

And this is equivalent to $\Bbb{R}P^2\# T^2$.

Euler Characteristics and Orientability:

Theorem: Two compact closed 2-manifolds $X$ and $Y$ are homeomorphic iff their orientability agree and $\chi(X)=\chi(Y)$ (equal euler characteristics).

Let $U$ be a small open nhbd around the central $S^1/{(x\sim -x)}$, and $V$ be an open nhbd of the complement of $U$ s.t. their intersection deformation retract onto $S^1$.

Then, Apply Mayer-Vietoris sequence, which will give you $$ H_p(X)\cong \begin{cases} \Bbb{Z} & p=0\\ \Bbb{Z}\oplus\Bbb{Z}\oplus\Bbb{Z}/2 & p=1\\ 0 & \text{otherwise} \end{cases} $$

So, by the definition of Euler characteristic in terms of betti number, we have $$\chi(X)=\sum_{p}(-1)^p\operatorname{rank}H_p(X)/T_p(X)=1-2=-1$$ where $T_p(X)$ denotes the torsion of $H_p(X)$. By the classification theorem of surfaces, we know that $X$ must be the connected sum of projective planes (must) and tori (may). Hence, we have the relationship $$\chi(X)=\chi(M_g)+\chi(N_k)-\chi(S^2)=2-2g+2-k-2=-1$$ for $k,g\in\Bbb{Z}_+\cup\{0\}$. Here, note that $M_g$ is the connected sum of $g$ tori, and $N_k$ is the connected sum of $k$ projective spaces.

There are two sets of solutions to this equation

  • $g=1$, $k=1$. This is $\Bbb{R}P^2\# T^2$
  • $g=0$, $k=3$. This is $\Bbb{R}P^2\# \Bbb{R}P^2\# \Bbb{R}P^2$

An it turns out that these two surfaces are homeomorphic, so we get the answer by the theorem stated before.

  • $\begingroup$ Thank u for the answer and you've took it right! But one thing that eludes me is how the third graph is obtained. I take it that the two same $c$ are put together and is then denoted as $x$, and $f=d^{-1}e$ is a $S^{1}$ in the second graph, but in the third one it connects one point (four vertices of the square identified as one) to another (one on $c$ or $x$), so what am I missing here? $\endgroup$
    – Juggler
    Nov 16, 2020 at 14:33
  • $\begingroup$ @Juggler The thing is that $c$ are identified, but $x$ is another edge along which we're about to cut the space into two pieces. I didn't draw $c$ because it connect the midpoints of two $f$. I hope it's clearer now. After we slice the space along $x$, we can glue $f$ together to obtain the last diagram. $\endgroup$
    – Kevin.S
    Nov 16, 2020 at 14:36
  • $\begingroup$ Yes, it’s all clear now. One more thing, the algebraic notation for labelling scheme you used, are there any references? Thx! $\endgroup$
    – Juggler
    Nov 16, 2020 at 14:45
  • $\begingroup$ @Juggler Yes. I learned that from Topology by Munkres in the chapter about classification theorem. But I might remember the wrong chapter as it’s been a while since my last revision on that book. $\endgroup$
    – Kevin.S
    Nov 16, 2020 at 14:48

Here is a third answer, to complement the two that Kevin has given above.

Imagine you have a cylinder $S^1\times I$. On the bottom $S^1\times 0$, you glue on a disk by identifying boundary circles. On the top $S^1\times 1$, you identify opposite points of the circle. What do you get? A projective plane! Because once you glue a disk on the bottom, what you have left is homeomorphic to a disk. Then you identify opposite points along the boundary circle, which is precisely how (in one approach) the projective plane is constructed.

Now let's go back in the other direction. Cut the bottom disk back off of your cylinder (so you've only glued opposite points on the top $S^1\times 1$). Now you have a projective plane with a disk removed. Gluing this to your torus with a disk removed (along boundary circles) is precisely how we construct the connected sum $\mathbb T\#\,\mathbb P.$

Now it remains to see that this is the same as directly gluing opposite points on your punctured torus. To see this, choose a bigger (closed) disk on your torus and drill out a small (open) disk from near its center. Now you have a cylinder/annulus, where one boundary circle is adjacent to the rest of the torus and the other boundary circle is undergoing this gluing of opposite points. Hence, we see that this is the cylinder-with-a-wonky-end glued onto the torus, from which we got $\mathbb T\#\,\mathbb P$ above. But only the smaller disk underwent any sort of change. So forgetting about the cylinder and the larger disk, this is the same as the construction that you initially described.


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