Join is simply connected

If $X$ is path-connected and $Y$ is arbitrary, then $X*Y$ is simply connected. 

 

$\textbf{Proof}.$

Define $U:=X*Y\setminus Y=X\times Y\times [0,1)/\sim,$ where $\sim: \ (x,y_1,0)\sim(x,y_2,0) $ for all $y_1,y_2\in Y.$  Thus $U$ is clearly open and deformation retracts onto $X\cong  X\times Y\times \{0\}/\sim.$ As $X$ is path-connected, so is $U$.

Fix $x_0\in X$. Let $V_1$ be the cone with vertex $x_0$ and base  $\{x_0\}\times Y\times\{1/2\}$. Also define $V_2:=X\times Y\times [1/2,1]/\sim$, where $\sim: \ (x_1,y,1)\sim(x_2,y,1) $ for all $x_1,x_2\in X.$ Set $V:=V_1\cup V_2$. Observe that $V_2$ deformation retracts onto  $\{x_0\}\times Y\times\{1/2\}$ (collapse everything to $Y$ and then push it to $\{x_0\}\times Y\times\{1/2\}$). But this is the base of the cone $V_1$, and $V_1$ deformation retracts to $x_0$. Hence $V$ is contractible.

Now argue as in the easier part of Seifert-van Kampen. The set $V$ is not open, but $U\cup \text{int}V=X*Y$. Thus every loop in $X*Y$ based at $x_0$ is a finite product of paths, each belonging to either $U$ or $V$. Moreover, since $X$ is path-connected, $U\cap V=V_1\cup (X\times Y\times [1/2,1))$ is path-connected. Thus every loop in $X*Y$ based at $x_0$ is homotopy equivalent to a finite product of loops at $x_0$, each belonging to either $U$ or $V$. Observe that $X$ is nullhomotopic in $X*Y$ (use a cone with base $X$ and vertex some point in $Y$), and hence $U$ is such as well. Thus the inclusion $U\hookrightarrow{} X*Y$ induces a trivial homomorphism $\pi_1(U)\to \pi_1(X*Y)$. Moreover, $\pi_1(V)$ is already trivial. Thus each loop in $X*Y$ is homotopy equivalent to a constant loop.