I would like to give examples of topological spaces such that their fundamental groups are, respectively, $\mathbb{Z} \oplus \mathbb{Z}_{n}$ and $\mathbb{Z} \ast \mathbb{Z}_{n}$. For the latter, I thought about this: let $X_{n}$ be the space obtained from $S^{1}$ by attaching a 2-cell with characteristic map $\chi (z) = z^{n}$. This map identifies all $n$-th roots of unity and can therefore be described as a quotient of $D^{2}$ with labels $a\ldots a$ ($n$ times). It's fundamental group, let alone, is $\langle a | a^{n} \rangle$, which is $\mathbb{Z}_{n}$ (in particular, if $n=2$, $X_{n} = P^{2}$). Now I thought about wedge-summing a copy of $S^{1}$ with this space. The resulting space would be, then, a space with fundamental group $\mathbb{Z} \ast \mathbb{Z}_{n}$, by Seifert Van Kampen's theorem. Is that right? Now, for the second part, I'm completely lost. If possible, I'd like to avoid using that the fundamental group of the cartesian product is the direct sum of fundamental groups.
[Math] How to determine space with a given fundamental group.
algebraic-topologyfundamental-groups
Related Solutions
The approach is correct, but you can use a simpler method. The space you care about is homotopic to the wedge sum of two spheres and one circle. Thus the fundamental group is Z. To see why this space is homotopic to the space I said, you just check first that you can contract the upper half-sphere, thus the lower half-sphere becomes to a sphere, and the torus becomes a sphere with its south and north poles glued together. A sphere with its south and north poles glued together is homotopic to a sphere with a line connect its south and north poles. After these deformations you have a space which is what I said above.
I'll give here another answer which I think is correct but I am not 100% sure. We will rely heavily on the first theorem at page 11 of Hatcher's Algebraic Topology, which basically allows us to do 2 things:
- We can kill any contractible line without changing homotopy type
- Instead of identifying two points we can just attach a $1$-cell at these two points.
Our first observation is that after identifying the edges, non of the vertices of the large square has been identifyied with eachother therefore we can collapse each edge of the edges $AB,BC,CD,DA$ to a point. The result space is Space $1$, where red points means identified (see the picture).
Now, we get the homotopy equivalent Space $2$ by replacing identifications by $1$-cells.
At last, we pull the outside point to the center, to get Space $3$, where the red lines are to be thought above the shape.
Now for the tricky part: The disc, without the red lines, deformation retracts to the wedge of $4$ circles. To picture this, imagine a strong gravity well at the center of the disk pulling everything radially at the center of the disc. Evertyhing either falls to the boundaries of the 4 existing holes or at the center of the disc, including the attaching points, therefore the red lines now became circles.
Therefore we end up wit the wedge of 8 circles and the fundamental group is $F_8$.
Best Answer
Your construction of a space with $\pi_1\cong\mathbb{Z}\ast\mathbb{Z}_n$ is good and the standard way to construct such a space.
The standard way to find an $X$ such that $\pi_1(X)\cong\mathbb{Z}\oplus\mathbb{Z}_n$ is, as you hinted at, to simply find a space $X_1$ with $\pi_1(X_1)\cong \mathbb{Z}$ and $X_2$ with $\pi_1(X_2)\cong \mathbb{Z}_n$ and then use the fact that $\pi_1(X_1\times X_2)\cong \pi_1(X_1)\times\pi_1(X_2)$. Using this approach, we find that the space $S^1\times (D^2\cup_{z^n}S^1)$ has the necessary fundamental group (here $(D^2\cup_{z^n}S^1)$ is the space you described in your question).
We can construct such a space though, without resorting to using the above theorem of product spaces. Note that the fundamental group of the torus $T^2$ is $\mathbb{Z}\times\mathbb{Z}$ which is generated by the meridinal and longitudinal loops on the torus. Call these representative elements $\alpha$ and $\beta$ respectively. Similarly to how you glued a disk to a circle to kill off certain elements in the fundamental group, we can do the same here. If we glue the boundary of $D^2$ to the image of the loop $\alpha$ via the map $f\colon z\mapsto z^n$ as before, then we introduce the relation $\alpha^n=0$ in the fundamental group, and as $\pi_1(T^2)$ is free abelian on the generators $\alpha,\beta$, there is no relation introduced on to the generator $\beta$. Hence $\pi_1(T^2\cup_fD^2)\cong \mathbb{Z}_n\oplus\mathbb{Z}$.
You can prove this more rigorously using Van-Kampen's theorem (take a small open neighbourhood on the torus around the loop $\alpha$, union the disk as the set $U$, and the torus minus the disk (and its boundary) as the set $V$).