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

Mathematics Asked on December 24, 2020

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 $S^1subset X$ unioned with the first 2 cell attached, and let $V$ be the same small open neighborhood of $S^1subset X$ unioned with the second 2-cell
attached. Then $Ucap V$ is that small open neighborhood of $S^1$, so we get that $U,V,Ucap V$ and are open and path connected, and $X=Ucup V$. 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 $mathbb{Z}/ nmathbb{Z}$ 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 $(i_{U’})_*(alpha)=nalpha$ because the attaching map maps $partial D^2$ $n
$
times around $S^1$. The interior of $D^2$ is contractible, so we get
$$pi_1(X’)=pi_1(U’)*_{pi_1(U’cap V’)}pi_1(V’)=mathbb{Z}*{0}/ nmathbb{Z}cong mathbb{Z}/nmathbb{Z}$$

The way I understand this result geometrically is that if I have a loop that wraps around $S^1subset X’$ 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 can“escape" (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 $S^1subset X$ 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 $mathbb{Z}/ mmathbb{Z}$ 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
$mathbb{Z}/2mathbb{Z}$. 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 $pi_1(Ucap V)congmathbb{Z}$. Let $i_U:Ucap Vhookrightarrow U$ and $i_V:Ucap Vhookrightarrow V$ be the corresponding inclusion maps. Then $(i_U)_*(1)= overline{1}in mathbb{Z}/2mathbb{Z}$ and $(i_V)_*(1)= tilde{1}in mathbb{Z}/3mathbb{Z}$.
SVK’s theorem gives
$$ pi_1(X)=pi_1(U)*_{pi_1(Ucap
V)}*pi_1(V)=langle overline 1,
tilde 1| overline 1 ^2=0, tilde 1
^3=0 , overline 1=tilde
1ranglecong 0 $$

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.

One Answer

Look at my answer to Can this counter example disprove the statement? The following is shown:

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 $pi_1(X) = 0$.

Answered by Paul Frost on December 24, 2020

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