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 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 $S^1\subset X$ 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.
 A: 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 $\pi_1(X) = 0$.
