Prove that the space $X = S^2 \cup [e_3, −e_3] \subset \Bbb R^3$ is not simply connected.

Question

Suppose we assume that $S^1$ is not simply connected. Let $e_i$ denote the standard basis vectors of $\mathbb{R}^3$. Prove that the space $X = S^2 \cup [e_3, −e_3] \subset \Bbb R^3$ is not simply connected.

We are looking for non-trivial homotopy class $[\gamma] \in \pi_1(X)$ right? First when it was stated that we know that $S^1$ is not simply connected I thought I could pick the equator $\gamma(t) = (\cos 2\pi t, \sin 2\pi t, 0)$, but I'm quite sure that this gives a non-trivial class.

So is the idea to consider a loop that encloses the line segment $[e_3, −e_3]$ like in the following picture:

it would seem that this gives a non-trivial class, but I don't see why I need the assumption on $S^1$ being not simply connected?

Answer 1

Often in algebraic topology it is easier to relate spaces to known ones and then rely on well-known results which may be non-trivial, in this case $\pi_{1}(S^1)$ and the Van Kampen Theorem.

The first comment above points the way. Since your resulting space $X$ has the homotopy type of $S^1\vee S^2$, by the Van Kampen Theorem it has $\pi_1(X)\cong\mathbb{Z}$.

That is probably easier than explicitly showing your red loop is essential even though geometrically it appears to be.

One way to think of $X$ as having the homotopy type of the above wedge is to view it as the quotient space of $S^2$ by identifying two distinct points such as the north and south poles. This is well-known and is covered in Hatcher's text and the link appears in this related question:

Fundamental group of $S^2$ with north and south pole identified

This source is also useful to help visualize this space:

https://www.math3ma.com/blog/clever-homotopy-equivalences

You could also use homology. If $A$ represents the red loop then the inclusion $A\rightarrow X$ should induce an injection (actually isomorphism) $H_1(A)\rightarrow H_1(X)$ and $A$ has the homotopy type of $S^1$.

Answer 2

You can use SvK in the following way. Hatcher tells us in theorem $1.26$ how to calculate the $\pi_1$ of a space $Y$ which results from attaching $2$-cells to the path-connected space $X$. I personally think the homotopy equivalence of your space with $S^1\vee S^2$ is quite hard to visualise and it implicitly uses the non-obvious (for me) fact that $D^2/([0,1]\times\{0\})$ is homeomorphic to $D^2$ again. That homeomorphism works in arbitrary dimension and relies on polar coordinates; $(r,\theta)$ is sent to the convex combination $((1-r)\cdot1+r\cdot\theta)$.

In the below I take the North pole to be a basepoint.

Here, we take $X$ to be $S^1$ with a connecting line from North to South; $\{0\}\times S^1\cup\{(0,0)\}\times[-e_3,e_3]$, if you want to geometrically fit everything in $\Bbb R^3$. By using SvK directly or by collapsing the contractible sub-CW-complex $\{(0,0)\}\times[-e_3,e_3]$ you can see that this space (obviously path-connected) has $\pi_1(X)\cong\Bbb Z\ast\Bbb Z$ with generators the loops $a$ and $b$ which are half-arcs from North down to South and then move along the connecting line $[-e_3,e_3]$ to get back to North.

Now we attach two copies of $D^2$ in the obvious way (gluing their $S^1$s to the $\{0\}\times S^1$ we had before by $x\mapsto(0,x))$ and we'll get $Y$ to be a copy of the space you care about. We can even write explicit geometric attaching maps $D^2\to S^2\cup[-e_3,e_3]$. According to theorem $1.26$, $\pi_1(Y)$ is isomorphic to $\Bbb Z\ast\Bbb Z$ where we quotient by the normal subgroup generated from the generating loops of $\{0\}\times S^1$; these loops are homotopic (modulo signs and inverses, it depends on how you setup and declare your orientations and your choice of $\pi_1(S^1)\cong\Bbb Z$) to $ab$ so we have $\pi_1(Y)\cong\langle a,b\,|\,ab=1\rangle$. In this group, $b=a^{-1}$ depends on $a$ and it's easy to realise $a\mapsto 1,\,b\mapsto-1$ defines a group isomorphism $\pi_1(Y)\cong\Bbb Z$.

Answer 3

Given the knowledge that $S^1$ is not simply connected, there is a simple proof that $X$ is not simply connected which does not use Van Kampen.

First, the space $X$ retracts onto the red closed curve in your image, which I'll denote $A \subset X$. The retraction map is simple to define: map each "constant latitude" circle to the unique point on $A$ that the circle intersects.

Second, for any retraction map $f : X \to A \subset X$ and for any $x \in A$, the induced homomorphism $f_* : \pi_1(X,a) \to \pi_1(A,a)$ is surjective.

Finally, $A$ is homeomorphic to the circle, and so $\pi_1(A,a) \approx \mathbb Z$. You can also notice that your closed curve $\gamma$ maps once around $A$ and so represents a generator of $\pi_1(A,a)$ (choose $a$ to be the north pole); it follows that $\gamma$ represents an infinite order element of $\pi_1(X,a)$.