how should I show that it is wedge of infinite circles?

Question

I know that the shape that we see it below is homotopy equivalent of wedge of infinite circles,so the fundamental group of it is $\prod _{1}(\vee _{\alpha \in A}S^{1})=\ast _{\alpha \in A }\mathbb{Z}$,but I don't know how should I show that it is wedge of infinite circles,also I can't imagine what is happening for this shape.please help me with your knowledge,thanks.

Answer 1

This is a difficult homotopy equivalence to visualize, but I've attempted to draw a picture of what's going on. First you poke a hole in your surface starting at $+\infty$ and pushing in from the right. Similarly poke a hole from the left. You can see this is a homotopy equivalence by analogy with an infinite cylinder, which can be visualized as having two dotted boundary components at $\pm\infty$. This transforms your surface into an infinite strip with a band connecting top and bottom and infinitely many tubes. Now look at the righ-hand side of my picture which shows a square with a tube in the middle. Once you draw in the indicated $1$-cells, the complement is a disk, which you can use to push the top boundary onto the remaining $1$-cells as indicated in the lower right. Do this for infinitely many squares in a row on your strip all at the same time, to get the bottom left picture. From here, it is easy to see this is a wedge of infinitely many circles. Note that you get one "extra" circle, besides the obvious infinitely many pairs.

Answer 2

Let $X$ be the infinite-holed torus and $G$ its fundamental group. You can write $X$ as an increasing union $X= \bigcup\limits_{n \geq 2} X_n$, where $X_n$ is a $4$-punctured $n$-holed torus; let $G_n$ denote its fundamental group, which is a free group of rank $2n+2$. In the figure below, $X_3$ is represented with a free basis of $G_3$ in red.

Now, you may notice that $G_{n+1}=G_n \ast \mathbb{F}_2$ (in particular, $G_n$ is naturally a subgroup of $G_{n+1}$) and that the inclusion $X_n \hookrightarrow X$ induces an injective homomorphism $G_n \hookrightarrow G$. Therefore, $\bigcup\limits_{n \geq 2} G_n$ is isomorphic to $\mathbb{F}_{\infty}$ and can be viewed as a subgroup of $G$. Because $\bigcup\limits_{n \geq 2} X_n = X$, in fact $G= \bigcup\limits_{n \geq 2} G_n$, and $G$ is a free group of (countably) infinite rank.

Added: 1) To show that $G_n \to G$ is injective, it is sufficient to show that if $c : [0,1] \to X_n$ is a loop such that $c=1$ in $G$ then $c=1$ in $G_n$. If $H : [0,1]^2 \to X$ is a homotopy between $c$ and the trivial loop, by compactness $\mathrm{Im}(H) \subset X_m$ for some $m \geq n$, hence $c=1$ in $G_m$. But $G_n$ is a subgroup of $G_m$ hence $c=1$ in $G_n$.

2) Thanks to the relation $G_{n+1}= G_n \ast \mathbb{F}_2$, there exist loops $a_1, a_2, \dots$ such that $\{a_1,\dots,a_{2+2n}\}$ is a free basis of $G_n$ for all $n \geq 2$. Therefore, $\{a_1,a_2, \dots\}$ is a free basis of $\bigcup\limits_{n \geq 2} G_n$.

Answer 3

Break up the space $X$ into infinitely countably many open pieces $\{X_i\}$, where each piece is homotopy equivalent to a torus minus two disjoint disks, and the intersection of each two consecutive pieces is homotopy equivalent to a circle. Pick a straight line $L$ that extends through all pieces, and pick a base point $y_0$ on this line. Let $Y_i$ be the union of $X_i$ and a small neighborhood of $L$ that deformation retracts onto $L$.

The open cover $\{Y_i\}$ of $X$ satisfies the conditions of the van Kampen theorem. The inclusion map $Y_i \cap Y_{i+1} \hookrightarrow Y_{i}$ maps the generator of $Y_i \cap Y_{i+1}$ to a generator of $Y_i$. The same is true for $Y_i \cap Y_{i+1} \hookrightarrow Y_{i+1}$. All other intersections have trivial fundamental groups. Since each $\pi_1(Y_i)$ is the free group on three generators, it follows that $\pi_1(X)$ is the free group on countably infinitely many generators.