Homotopy equivalence of $S^{2} \vee S^1$ to $S^{2} \cup A$ where A is a line segment joining noth and south poles I have some problems trying to show homotopy equivalence of $S^{2} \vee S^1$(one-point union) to $S^{2} \cup A$ where $A$ is a line segment joining north and south poles of a sphere. I understand the general flow of the argument but can't actually construct the maps between the two spaces.
This question can be found in Bredon's Topology and Geometry Book Chapter 14 Exercise 1 
Thanks! 
 A: There are two approaches 


*

*If $(X,B)$ is a cofibered pair of spaces and $B$ is contractible, then the quotient map $X\to X/B$ is a homotopy equivalence. In your case, you could take $X=S^2\cup A$ and $B=\{(x,y,z)\in S^2\mid x\ge0=y\}$. Since $S^2/B$ is homeomorphic to a $2$-sphere (which can be cumbersome to show), and $A/(A\cap B)$ is homeomorphic to a circle, $X/B$ is homeomorphic to $S^2$ with a circle attached to it at the point $B$, and the quotient map $X\to X/B$ is the desired homotopy equivalence.

*The second approach presents a less difficult way: We regard $X$ as the sphere $S^2$ with an interval $A=[0,1]$ attached to it via the map $f:B=\{0,1\}\to S^2$ sending $0$ to the south pole and $1$ to the north pole. So $X$ is the pushout of the diagram $A\hookleftarrow B\xrightarrow f S^2$, written as $A\cup_f X$. Now $f$ is homotopic to the map $f_1:B\to\{(1,0,0)\}$, and since $(A,B)$ is cofibered, we can use the general fact that homotopic attaching maps $f\simeq f_1:B\to S^2$ give rise to homotopy equivalent pushouts $A\cup_f S^2$ and $A\cup_{f_1}S^2$, the later being $S^2\vee S^1$.


I think you can find the proofs for both facts in Hatcher's Algebraic Topology, chapter 0. The proof for the second fact, also gives an idea of the homotopy equivalence implied by the setting.
