$\mathbb{S}^2/{\sim} \simeq \mathbb{S}^2\vee\mathbb{S}^1$ where $(0,0,1)\sim (0,0,-1)$

Question

I am trying to understand the example $0.8$ of Hatcher's book (pag $11$), I only know singular homology and I do not know anything about CW-complex, ​​so it is difficult to interpret what it says there. How can I get to that $\mathbb{S}^2/{\sim} \simeq \mathbb{S}^2\vee\mathbb{S}^1$ where $(0,0,1)\sim (0,0,-1)$ without using CW-complex structures, only with basic knowledge of singular homology?

Thank you!

Answer 1

I don’t think this example refers to any background in homology. The point is, by introducing the intermediate picture of $X$, it is clear $X/A$ and $X/B$ are both homotopy equivalent to $X$. Thusly $X/A$ and $X/B$ are homotopy equivalent. Since $X/A$ is $S^1/\sim$, where the relation is north and south poles being glued together, and $X/B$ is $S^2 \wedge S^1$, they are homotopy equivalent.

Sorry for the format, but I am writing on phone.

Answer 2

Here is a sketch of a more direct proof. As in Hatcher's argument, we just need to show the quotient maps $X\to X/A$ and $X\to X/B$ are homotopy equivalences, to get a homotopy equivalence between $X/A\cong S^2/S^0$ and $X/B\cong S^2\vee S^1$. I will sketch this for the quotient map $f:X\to X/A$; the map $X\to X/B$ works similarly.

To show $f$ is a homotopy equivalence, we need to find a homotopy inverse. We can map $X/A$ to $X$ by collapsing each of the two vertical circles in the picture of $X/A$ to a point and then similarly collapsing each vertical cross-section to the right of them in the picture to two points (one on top and one on bottom). This turns the entire right half of the picture into an arc, which we can then map to the arc $A$ in $X$, This thus defines a map $g:X/A\to X$.

To show $g$ is a homotopy inverse of $f$, we must show $gf$ and $fg$ are homotopic to the identity. What does $gf$ do? It collapses $A$ to a point, and then shrinks bits of the sphere that were around the endpoints of $A$ to arcs and maps them to $A$ (with $A$ itself all mapping to the midpoint of $A$), stretching out the rest of the sphere to cover the whole sphere. Intuitively, this is homotopic to the identity because we can start with the identity and gradually shrink $A$ down towards the midpoint, while stretching more and more of the sphere surrounding its endpoints to map to the rest of $A$ and stretching the rest of the sphere to fill out the sphere.

How about $fg$? It collapses the right half of $X/A$ down to an arc, and then shrinks that arc back down to a single point and stretches the left half of $X/A$ to cover all of $X/A$. Again, this is homotopic to the identity because you can do this process gradually, shrinking more and more of the right half to a point and gradually stretching the left half to cover more and more of the space.

Now of course, none of this is a rigorous proof; I've just given rough geometric descriptions of all the maps involved. Actually writing them down explicitly can be done but is quite a pain. This is the value of the basic facts about CW complexes that Hatcher is using. In particular, using the general result of Proposition 0.17 and some general facts about extending homotopies over CW complexes (specifically, Proposition 0.16), he can conclude that $f$ is a homotopy equivalence without needing to find all the relevant maps explicitly, using just the fact that $A$ is contractible and is a subcomplex of $X$.

Answer 3

I agree with the previous answer that this is nothing to do with homology. It is about the homotopy type of adjunction spaces.

7.5.5 (Corollary 1) of Topology and Groupoids says that if $(X,A)$ is cofibred and $f \simeq g: A \to B$ then there is a homotopy equivalence of adjunction spaces $B \cup _f X \simeq B \cup_g X \; \text{rel}\; B$.

So in your picture the space $X$ is obtained by attaching an internal $I= [0,1]$ joining the North and South poles and in what is called $X/B$ is also obtained by mapping $0,1$ to the South pole.