Attaching $2$-cell to a circle

I get this problem from my past Qual.

" For $$n=1,2,\dots$$ let $$f_n\colon S^1\to S^1$$ be the map $$z\mapsto z^n$$. Define $$X(n)=S^1\cup_{f_n} e^2$$ as the cell complex obtained by attaching a 2-cell to the circle via the attaching map $$f_n$$. Similarly, for $$m,n=1,2,\dots$$ define $$X(m,n)=S^1\cup_{f_m} e^2_1\cup_{f_n}e^2_2$$ as the cell complex obtained from the circle by attaching two 2-cells, with attaching maps $$f_m$$ and $$f_n$$.

a) Show that if $$m>n$$ then the maps $$f_m,f_{m-n}\colon S^1\to S^1\hookrightarrow X(n)$$ are homotopic as maps into $$X(n)$$.

b) Show that for all $$m,n$$ the space $$X(m,n)$$ is homotopy equivalent to the wedge sum $$S^2\vee X(k)$$ for a certain $$k=k(m,n)$$. Give an explicit formula for $$k$$ in terms of $$m,n$$."

For $$(a)$$, I consider the homotopy $$H\colon S^1\times I \to S^1$$ where $$H(z,t)=z^{m-nt}$$. This is clearly continuous. Hence the homotopy $$S^1\times I \to S^1 \to X(n)$$ is our desired homotopy. So we're done.

The problem is, I don't even care about $$X(n)$$, so it must be wrong somewhere. Can you help me with this?

I have no clue how to do $$(b)$$ since I cannot visualize $$X(n),$$ let alone $$X(m,n)$$.

• What is $z^{3/2}$? Or $z^{4/3}$? Or $z^{\pi}$? Over complex numbers of course. Your $H$ is not even well defined. The concept of $z^n$ is clear when $n\in\mathbb{Z}$, but not so much when $n\in\mathbb{R}$. And when you rewrite complex exponantiation as $e$-power, then it boils down to choosing complex logarithm branches. But there is no such choice making $H$ continuous. Otherwise $S^1$ would be contractible. So "clearly" (pun intended) $H$ is either ill-defined or discontinuous. – freakish Feb 7 at 14:56
• For help "visualizing" $X(n)$, notice that $X(2)\cong \mathbb{R}P^2$; saying "it's a disk attached to a circle by a map of degree $n$" is as clear as it gets in this case. As a hint for how to do a), notice that $\pi_1(X(n))\cong \mathbb{Z}/n\mathbb{Z}$, generated by $f_1$ (prove using Seifert-van Kampen). The fundamental group of $X(n,m)$ similarly is generated by $f_1$, but in this group there are the two relations $f_1^n = 0$ and $f_1^m = 0$; can you describe this group? – William Feb 7 at 15:09

I think this is a good Algebraic Topology question, because if you approach it a certain way then the algebra does a lot of the heavy lifting but there are still one or two crucial topological inputs.

Hint for a):

Use Seifert-van Kampen to compute $$\pi_1(X(n))$$, and in particular see that $$f_1$$ is a generator. Notice that $$f_m = f_1^m \sim *^m f_1$$ (where $$*$$ is path concatenation in $$\pi_1$$).

Hint for b):

As a warm-up first prove the following two equivalences: $$X(n,n)$$ is homotopy equivalent to $$X(n,0)$$ (where $$f_0$$ is the constant map) and $$X(n,0)$$ is homeomorphic to $$X(n) \vee S^2$$. For $$n< m$$ consider part a) and the Euclidean algorithm.

More thorough hints for b):

Instead of directly writing down an explicit homotopy equivalence between $$X(n,m)$$ and $$X(k)\vee S^2$$ we can use the following key observation, which is the main topological input: if $$f_m,f_{m'}\colon S^1 \to S^1 \hookrightarrow X(n)$$ are homotopic then $$X(n, m)$$ is homotopy equivalent to $$X(n,m')$$ (see for example this question). This fact together with part a) implies that if $$n < m$$ then $$X(n, m)\simeq X(n, m-n)$$. Run the Euclidean algorithm on $$n$$ and $$m$$.

Observe that we could a priori determine what $$k$$ would have to be by assuming the homotopy equivalence in part b) is true and computing $$\pi_1(X(n,m))$$. Applying van Kampen to $$X(n) \cup_{f_m} e^2$$ we find $$\pi_1(X(n, m))$$ is again generated by $$f_1$$, but we are now given two relations: $$n[f_1] = 0$$ and $$m[f_1] = 0$$. We still know the group is cyclic and has torsion, so it must be isomorphic to $$\mathbb{Z}/k$$ for some $$k$$: determine $$k$$, with the help of the Euclidean algorithm.