# Attaching a disk $D^2$ along the boundary circle to a circle $S^1.$

Let $$Y$$ be the space obtained by attaching a disk $$D^2$$ along the boundary circle to a circle $$S^1$$ by a map that wraps the boundary circle around the other circle 3 times, i.e., the following square is a pushout. Calculate $$\pi_{1}(Y).$$

My questions are:

1- I do not understand the statement: "by attaching a disk $$D^2$$ along the boundary circle" what do the question mean by $$along the boundary$$? does it mean tangentially? Also, are there other ways of attaching a disk?

2- I feel like I should use Van Kampen theorem but I do not know how to divide my space $$Y$$ into union path-connected open sets each containing the basepoint $$y_{0} \in Y$$?

Suppose we have a closed disc $$\bar{D}^2$$ whose boundary $$\partial D^2$$ is attached to a circle $$S^1$$ by a map $$\gamma:\partial D^2\to S^1$$ that wraps $$\partial D^2$$ in total $$n$$ times around $$S^1$$. We call the resulting space $$Y_n$$.

Let $$U$$ and $$V$$ be open subsets of $$Y_n$$ defined as follows. The set $$U$$ is given by $$(U\cap D^2)\cup S^1$$, where $$D^2$$ is the interior of $$\bar{D}^2$$, and $$U\cap D^2$$ is a narrow strip on the outer edge of $$D^2$$ (so that $$\partial(U\cup D^2)$$ contains $$\partial D^2$$). The set $$V$$ is just $$D^2$$. My awful picture may help explain this. The space $$Y_n$$ (on the left, where the orange arrows denote the attaching map $$\gamma$$) is the union of $$U$$ (the yellow subset) and $$V$$ (the pink subset).

Note that each of $$U$$ and $$U\cap V$$ has a deformation retract to just $$S^1$$, but $$V$$ is contractible. That is $$\pi_1(U)\cong\Bbb Z$$, $$\pi_1(V)\cong\{1\}$$, and $$\pi_1(U\cap V)\cong \Bbb Z$$. Now observe that $$\pi_1(U\cap V)\to \pi_1(U)$$ is given by multiplication by $$n$$ because each simple loop in $$U\cap V$$ wraps $$n$$ times around $$S^1$$ (and $$\pi(U\cap V)\to \pi_1(V)$$ is trivial). By van Kampen's theorem, $$\pi_1(Y_n)=\pi_1(U)\underset{\pi_1(U\cap V)}{*}\pi_1(V)\cong (\Bbb Z*\{1\})/(n\Bbb Z)\cong \Bbb Z/n\Bbb Z.$$ Indeed, we can see that $$\pi_1(Y_n)$$ is generated by a generator $$g$$ of $$\pi_1(S^1)$$. When you have a loop homotopic to $$ng$$, it is homotopic to $$\partial D^2$$, and can then be contracted along $$\bar{D}^2$$ to a point.

• I do not understand your last 2 lines ..... could you please explain them ? – Secretly Oct 7 at 11:48
• So basically you have $\bar{D}^2$ attached to $S^1$ by $\gamma$ right? Well, you can take $g=[S^1]$ itself. Then $g$ generates $\pi_1(Y_n)\cong \Bbb Z/n\Bbb Z$. In other words, $ng$ as an element of $\pi_1(Y_n)$ is the identity $0$ of $\Bbb Z/n\Bbb Z$. To see this, if you have an $n$-fold loop $ng$, then it is precisely the boundary of $\bar{D}^2$ (recall that $\partial D^2$ wraps $n$ times around $S^1$). But the boundary of $\bar{D}^2$ can be contracted to a point on $\bar{D}^2$. – WE Tutorial School Oct 7 at 16:49
• math.stackexchange.com/questions/3426826/… Could you please look at this question if you have time? – Secretly Nov 8 at 5:59

Let's start from the begining. $$S^1$$ is given and another, distinct $$D^2$$ is given. The boundary $$\partial D^2$$ of $$D^2$$ is $$S^1$$ as well, but since it is distinct I will denote it as $$\partial D^2$$.

1- I do not understand the statement: "by attaching a disk $$D^2$$ along the boundary circle" what do the question mean by $$along the boundary$$? does it mean tangentially?

The concept is the same as in CW complex construction. You start with a map $$f:\partial D^2\to S^1$$ (in your case the triple winding) and then you glue $$D^2$$ and $$S^1$$ along this map, i.e. you take the quotient space

$$(D^2\sqcup S^1)/\sim$$

where "$$\sim$$" is generated by $$x\sim f(x)$$ for $$x\in\partial D^2$$. In particular note that if $$f(x)=f(y)$$ then $$x\sim y$$.

Also, are there other ways of attaching a disk?

Of course. If you glue along say identity $$f(x)=x$$ then the result is simply $$D^2$$. The same goes for the antipodal map $$f(x)=-x$$. But in your case this is something different. Note that if you attach along a double winding you obtain the real projective space $$\mathbb{R}P^2$$.

2- I feel like I should use Van Kampen theorem but I do not know how to divide my space $$Y$$ into union path-connected open sets each containing the basepoint $$y_{0} \in Y$$?

So let's generalize this a bit and assume that the attaching map winds $$n$$ times. Calculating the fundamental group for general $$n$$ is very similar to calculating it for $$\mathbb{R}P^2$$. Here is the answer that goes through the process in details: An intuitive idea about fundamental group of $\mathbb{RP}^2$ The core idea there is that they use the path lifting property of coverings instead of Van Kampen.

Try to generalize it (the quotient is no longer $$x\sim -x$$ but $$x$$ is now related to $$n-1$$ other points on $$\partial D^2$$) and note that the result should be $$\mathbb{Z}_n$$.

• Why we will not use Van Kampen? if we used it will we have intersection of 3 sets instead of 2? – Secretly Oct 4 at 8:37
• @hopefully well, I've tried to apply Van Kampen directly but it didn't lead me to the solution. I might have missed something though. I did not want to say that Van Kampen is a wrong way, I've reformulated that sentence. – freakish Oct 4 at 9:18