$\pi_2$ of a 2-cell attached to a circle via degree $m$ map

Question

Consider $X = D^2 \cup_f S^1$ where $f : S^1 \to S^1$ is a degree $m$ map. How may we compute $\pi_2(X)$?

Answer 1

Start by having a look at the theory of crossed modules (say in the paper by Brown and Huebschmann: (R. Brown and J. Huebschmann, 1982, Identities among relations, in R. Brown and T. L. Thickstun, eds., Low Dimensional Topology, London Math. Soc Lecture Notes). Another source might be Loday's article on homotopical syzygies, in Une d ́egustation topologique: Homotopy theory in the Swiss Alps, volume 265 of Contemporary Mathematics, 99 – 127, AMS. The point is that the fundmental group of your space is clear (from van Kampen's theorem) and it is given by a group presentation. As X is 2-dimensional, its $\pi_2$ is the module of identities of that presentation ... over to you!

Answer 2

One can be more ambitious and try to compute the homotopy type, of which the homotopy groups, even as $\pi_1$-modules, are often pale shadows. For an introduction to the ideas, see this paper Modelling and Computing Homotopy Types:I, to appear in Indagationes Math. in 2017 as part of a special issue in honor of L.E.J. Brouwer.

To add to the above, one can generalise the given problem to ask for the homotopy $2$-type of the mapping cone $C(B(f))$ where $f: G \to H$ is a morphism of groups, and $B(G)$ is the classifying space. This problem is dealt with in Chapter 5 of the book partially titled Nonabelian Algebraic Topology (EMS Tract in Mathematics vol 15, (2011)). It is a consequence of a 2-d Seifert-van Kampen Theorem that the homotopy 2-type is determined by the "induced crossed module" $f_*(G) \to H$; a number of specific calculations are given in that Chapter. In the case $G=H=C$, the infinite cyclic group with generator $x$ and $f(x)= x^n$, then $f_*(G)\cong \mathbb Z[C_n]$ generated as a $C$-module by $x_1$, say, and the crossed module is $\delta_2: \mathbb Z [C_n] \to C$ where $\delta_2(x_1) = x^n$ (see Example 10.2.4 of the above book). If $G,H$ are finite then so also is $f_*(G)$.

Answer 3

Big hint: Letting $\tilde X \to X$ be the universal covering space, the exact sequence of a fibration implies that $\pi_2(\tilde X) \approx \pi_2(X)$. Since $\tilde X$ is simply connected, it follows from the Hurewicz theorem that $\pi_2(\tilde X) \approx H_2(\tilde X;\mathbb{Z})$. So you just have to compute $H_2(\tilde X;\mathbb{Z})$.

You can use Van Kampen's theorem to compute that $\pi_1(X) \approx \mathbb{Z} / m \mathbb{Z}$, which should help you to construct the universal cover $\tilde X$ and compute its second homology.

Answer 4

Consider the group $G=\langle a \mid a^n=1\rangle$. We should be able to see that the infinite cyclic group corresponds to $\pi_1(S^1)$, and we can start with this. To actually get the presentation, we need to add a relation. One way to do this, is to take an $n$-gon, on which $G$ acts transitively (by moving one vertex counter-clockwise.)Fix a basepoint, on $S^1$, and make the identification that each vertex on the $n$-gon corresponds to it (this is essentially saying that the polygon with the appropriate orientation is the Cayley graph.)

Then the map you describe gives a space with fundamental group $G$.

For $\mathbb Z/2 \mathbb Z$ the universal cover will be $\mathbb R P^2$. This is essentially done by taking the copy of two disks, and identifying them via an antipodal action on the boundary. For $\mathbb Z/n \mathbb Z$ it will be $n$ disks, all being identified on their boundary. The group $G$ acts on this space by cyclic permutation of each ddisk in the stack, followed by a $2\pi/n$ rotation. The $1$-skeleton is still the cayley graph for $G$, but the $\pi_2$ component should just be $\mathbb Z^{n-1}$.

To see that this is enough, note that $\pi_1$ vanishes for the cover, so $\pi_2(\tilde{X}) \cong \pi_2(X)$. Furthermore, we can argue that it is torsion free by algebraic arguments, (the Hurewicz map reference by Lee Mosher will do, since the cover is $1$-connected) or by wishful thinking :), but I think that this is the idea.