Let $X$ and $Y$ are copies of $S^2$ taking points $x_1,x_2 \in X$ and $y_1, y_2 \in Y$ then $\pi_1(X \bigsqcup Y/x_1\sim y_1,y_2\sim x_2 )?$

Question

Recently I have been reading Hatcher's algebraic topology book; in chapter zero, there is a problem that if take the sphere of dimension $2$ (symbolically $S^2$)and identify two points on $S^2$ then it is homotopy equivalent to $S^2\vee S^1$. Now I have a question here if we take two identical copies of $S^2$, call it $X$ and $Y$; I am taking two different points from $X$, say $x_1~and ~x_2 \in X$ and, taking two points from $Y$ say $y_1~ and ~y_2 \in Y$. Let $Z$ be the quotient space defined by identifying the points $x_i$ with $y_i$ for $i=1,2$ i.e $Z=X \bigsqcup Y/x_1\sim y_1,y_2\sim x_2 $, is $Z$ homotopy equivalent to $S^2 \vee S^1 \vee S^2$? Actually, I want to find the first fundamental group of $Z$.

Answer 1

Yes, $Z\space{\sim}{\space}S^2\space\vee{\space}S^1{\space}{\vee}{\space}S^2$ by homotopy equivalence. You can think of the equivalence working like this picture, where the two spheres are retracting along sides of the circle.

Since the homotopy equivalence of spaces yields an isomorphisms between fundamental groups, you have $\pi_1(Z)\cong\pi_1(S^2\space\vee{S^1}\space\vee{S^2})\cong\pi_1(S^2)\ast\pi_1(S^1)\ast\pi_1(S^2)\cong\pi_1(S^1)\cong\mathbb{Z}$, since fundamental group of $S^2$ is trivial, and fundamental group of a wedge sum of space is isomorphic to free products of fundamental groups.