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The question below is from an old topology qualifying exam. I am mostly stuck on parts (c) and (d).

Let $X$ be a 2-dimensional torus $T^2$ with the interior of a small disk $D \subset T^2$ removed (this space is also called a handle).

(a) Prove that $X$ is homotopically equivalent to a bouquet of two circles $S^1 \vee S^1$. To do this, it’s convenient to represent $T^2$ as a unit square with the opposite sides identified.

I am fine with part (a) - I understand that if you puncture the torus, it deformation retracts to $S^1 \vee S^1$.

(b) Compute the fundamental group $\pi_1(X)$ of $X$ and present its generators explicitly as loops in $X$.

And for part (b), I can use Van Kampen on $S^1 \vee S^1$ - however, I'm confused by the wording, and wonder if $\pi_1(X) = \langle a\rangle * \langle b\rangle$ is the correct answer.

(c) Explicitly express the class of $\partial X = \partial D$ in $\pi_1(X)$ in terms of the generators from part (b).

I tried to find the boundary in $\mathbb{R}^2$ using the definition:

$$ \partial A = \overline{A} \cap \overline{(\mathbb{R}^2 - A)}$$

and the identification of $X$ as a square with a disc removed from part (a). I found that $\overline{T^2 - D}$ is $T^2 - D$ plus the boundary of $D$ (the picture on the left), and $\overline{(\mathbb{R}^2 - A)}$ is everything outside the square (including the boundary)[picture on the right]. So if I intersect them, I get that the boundary would be a circle inside a square (where both are not filled in)[the bottom picture]. enter image description here Am I thinking of that correctly? I have no idea what the fundamental group of this would be (I can only guess it's $S^1 \vee S^1$ plus the circle on its own, which doesn't make sense to me. And by class, does that mean all the elements of the fundamental group, $\pi_1(X)$, that are along the boundary of $X$?

(d) Prove that $\partial X$ is not a retract of $X$. Hint: What is $\pi_1(\partial X)$?

Visually, it makes sense that the boundary can't retract to $X$ if I'm thinking of $\partial X$ correctly. I think for this part the best thing would be to find $\pi_1(X)$, but I can't without part (c).

I apologize for the atrocious pictures. Feel free to draw bad pictures to help with an answer!

dorkichar
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    Good work on the earlier parts! I think your calculation for the boundary isn't quite right - $\Bbb R^2$ isn't the ambient space here. You want the boundary of $X$ in $T^2$, which as the question says is the same as the boundary of $D$ - it's the circle on the edge of $D$. Then (c) wants you to identify this circle with a loop and write that loop in terms of the generators you have for $\pi_1(X)$ (up to homotopy). You can also think of this part as "explicitly calculate the map induced on fundamental groups by the inclusion of $\partial X$ into $X$", which will help for (d)... – Izaak van Dongen Jul 30 '22 at 00:00
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    For part b) a better way (perhaps) to write the group presentation is probably $\pi_1(X)=\langle a,b\rangle$. For part c), an intuitive way is to think about which loop is that boundary circle homotopic to (as Izaak mentioned, your calculation isn't quite right for that part). – Kevin.S Jul 30 '22 at 03:07
  • My first thought for part c) is that the boundary is homotopic to the square above, so we have $aba^{-1}b^{-1}=1$ representing going around the circle one time using the generators from part b). And since $\pi_1(\partial X) \cong \pi_1(S^1) = \mathbb{Z}$, we can think of going around the circle $n$ times as $(aba^{-1}b^{1})^n = 1+1+\ldots +1 = n$. Does this make sense, or am I going in the wrong direction? And since $\pi_1(\partial X) = \mathbb{Z} \neq \pi_1(X) = <a,b>$, $\partial X$ is not a retract of $X$. – dorkichar Jul 30 '22 at 17:01
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    Yes, though I don't know exactly what you're calling $a$ and $b$, the loop being in the class of $aba^{-1}b^{-1}$ looks right! I'm afraid your argument for (d) isn't right though - retracts don't always have the same fundamental group (that would be deformation retractions). If $r: X \to \partial X$ is a retract and $\iota: \partial X \to X$ is the inclusion, then you get an equality of induced homomorphisms on fundamental groups: $r_* \circ \iota_* = \mathrm{Id}_{\Bbb Z}$. You should argue that this is absurd (using what you know from (c)!) – Izaak van Dongen Jul 30 '22 at 20:14
  • @Kevin.S, why not let OP have a go at figuring it out? Besides, I don't think the inclusion does induce the trivial map - it sends the generating loop to $aba^{-1}b^{-1}$ (or something similar), which is not the identity element of the free group on $a$ and $b$. – Izaak van Dongen Jul 31 '22 at 08:45
  • @IzaakvanDongen Oh, indeed, it's non-trivial; I thought it was a torus. – Kevin.S Jul 31 '22 at 10:51

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