# Calculating the Fundamental Group of a CW Complex with Attaching Maps of Varying Degrees

I'm reviewing old homework questions for an upcoming topology exam, and I came across a question that I did not fully understand at the time. In the solution to this homework question, I had to compute the fundamental group of the following $$CW$$ complex. $$X$$ is the CW complex obtained from $$S^1$$ with its usual cell structure by attaching two 2-cells by maps of degrees 2 and 3 respectively.

I approached the problem as follows. Let $$U$$ be an small open neighborhood of $$S^1\subset X$$ unioned with the first 2 cell attached, and let $$V$$ be the same small open neighborhood of $$S^1\subset X$$ unioned with the second 2-cell attached. Then $$U\cap V$$ is that small open neighborhood of $$S^1$$, so we get that $$U,V,U\cap V$$ and are open and path connected, and $$X=U\cup V$$. I wish to apply SVK's theorem to $$U$$ and $$V$$ to calculate the fundamental group of $$X$$. However, to find $$\pi_1(U)$$ and $$\pi_1(V)$$, I think I have to apply SVK's theorem to each first. In order to avoid duplicate work, I will prove that the fundamental group of CW complex $$X'$$ obtained from $$S^1$$ by attaching a two cell with maps of degree $$n$$ isomorphic to $$\mathbb{Z}/ n\mathbb{Z}$$ and apply it separately to $$U$$ and $$V$$ which are special cases. If one lets $$U'$$ be some small open neighborhood of $$S^1$$ (and hence it contains $$\partial D^2$$) and let $$V'$$ be the interior of the attached 2-cell, $$U'\cap V'$$ is connected and retracts to $$S^1$$ so it has fundamental group isomorphic to $$\mathbb{Z}$$. Let $$i_{U'}:U'\cap V' \hookrightarrow U'$$ be the inclusion map and $$\alpha$$ be a generator for $$\pi_1(U'\cap V')$$. Then $$(i_{U'})_*(\alpha)=n\alpha$$ because the attaching map maps $$\partial D^2$$ $$n$$ times around $$S^1$$. The interior of $$D^2$$ is contractible, so we get $$\pi_1(X')=\pi_1(U')*_{\pi_1(U'\cap V')}\pi_1(V')=\mathbb{Z}*\{0\}/ n\mathbb{Z}\cong \mathbb{Z}/n\mathbb{Z}$$

The way I understand this result geometrically is that if I have a loop that wraps around $$S^1\subset X'$$ once, I cannot contract it to a constant loop because it wraps around $$S^1$$ and $$S^1$$ is not contractible. However, if I wrap around $$S^1$$ $$n$$ times, then the loop traverses the boundary of the attached 2 cell $$D^2$$ and canescape" (via homotopy) to the 2-cell $$D^2$$. $$D^2$$ is contractible of course, so this loop contracts to the trivial loop.

Back to the original question. My geometric arguments seems to give the fundemantal group of $$X$$ almost immediately. Indeed, if I have a loop that wraps around $$S^1\subset X$$ once, it is not trivial (like before). However, if I wrap around $$S^1$$ twice, then my loop can "escape"to the 2-cell attached with the map of degree 2. Thus if a loop wraps around $$S^1$$ 3 times, then the loop first wrapping arounds twice, which is homotopic to the identity, so it is homotopic to a loop that wraps around once. Thus attaching a 2-cell with a map of higher degree adds NOTHING to the fundamental group. More generally, if I attach 2-cells to $$S^1$$ with maps of varying degree, the fundamental group is simply isomorphic to $$\mathbb{Z}/ m\mathbb{Z}$$ where $$m$$ is the lowest degree of the attaching maps. Is this intuition correct? This reasoning leads me to believe that the fundamental group of $$X$$ is simply $$\mathbb{Z}/2\mathbb{Z}$$. This seems like a fairly rigorous argument to me, but I would like to calculate directly using SVK's theorem.

When I intersect $$U$$ and $$V$$, I will get a space that contracts to the $$S^1$$ (the 1-skeleton). Let $$1$$ be the generator for $$\pi_1(U\cap V)\cong\mathbb{Z}$$. Let $$i_U:U\cap V\hookrightarrow U$$ and $$i_V:U\cap V\hookrightarrow V$$ be the corresponding inclusion maps. Then $$(i_U)_*(1)= \overline{1}\in \mathbb{Z}/2\mathbb{Z}$$ and $$(i_V)_*(1)= \tilde{1}\in \mathbb{Z}/3\mathbb{Z}$$. SVK's theorem gives $$\pi_1(X)=\pi_1(U)*_{\pi_1(U\cap V)}*\pi_1(V)=\langle \overline 1, \tilde 1| \overline 1 ^2=0, \tilde 1 ^3=0 , \overline 1=\tilde 1\rangle\cong 0$$ So, there seems to be a problem with my computation. How might I fix my computation? Is there a different (perhaps shorter) way of calculating $$\pi_1(X)$$? Originally I was thinking about removing a point in the interior of each attached 2-cell and proceeding similarly to the way one would show the fundamental group of the sphere is trivial using SVK's theorem.

• $\pi_1(X)$ is the free group with one generator $z$, and two relations $z^2=e$ and $z^3=e$, therefore $\pi_1(X)$ is trivial. Commented Dec 22, 2020 at 20:22
• Yes, thank you for pointing out that typo. I corrected it to $\overline{1}=\tilde{1}$.
– MEG
Commented Dec 22, 2020 at 20:47
• However, this still gives the trivial group. There must be a error in my work.
– MEG
Commented Dec 22, 2020 at 20:54

Let $$X$$ be any space. Take an element $$g \in \pi_1(X,x_0)$$, represent it by a map $$\gamma : S^1 \to X$$ and attach a $$2$$-cell to $$X$$ via $$\gamma$$. If $$i : X \to X' = X \cup_\gamma D^2$$ denotes inclusion, then $$i_* : \pi_1(X,x_0) \to \pi_1(X', x_0)$$ is an epimorphism whose kernel is the normal subgroup $$N(g)$$ of $$\pi_1(X,x_0)$$ generated by $$g$$.
Now let $$f_k : S^1 \to S^1$$ be a map of degree $$k$$ (we may take $$f_k(z) = z^k$$). Consider $$X_2 = S^1 \cup_{f_2} D^2$$ with inclusion $$i : S^1 \to X_2$$ and $$X = X_2 \cup_{if_3} D^2$$ with inclusion $$j : X_2 \to X$$. The space $$X$$ is the CW complex of your question. Let $$g = [f_1]$$ be the canonical generator of $$\pi_1(S^1) \approx \mathbb Z$$. By the above result $$i_* : \pi_1(S^1) \to \pi_1(X_2)$$ is an epimorphism whose kernel is generated by $$[f_2] = 2g$$. Thus $$i_*(g)$$ is the generator of $$\pi_1(X_2) \approx \mathbb Z_2$$. The map $$if_3 : S^1 \to X_2$$ represents the homotopy class $$i_*([f_3]) = i_*(3g) = 3i_*(g) = i_*(g)$$. But $$j_* : \pi_1(X_2) \to \pi_1(X)$$ is an epimorphism whose kernel is generated by $$i_*(g)$$. Therefore $$\pi_1(X) = 0$$.
• @blueskyscroll Why do think that $X$ is homotopy equivalent to $S^2$? Anyway, it is easy to see that $X$ is not homeomorphic to $S^2$. Removing a point from $S^2$ gives a contractible space (homeomorphic to $\mathbb R^n$). But removing an interior point of the second attached $D^2$ gives a space contaning $X_2$ as a strong deformation retract, and this is not contractible. Commented Apr 5, 2021 at 8:43