Property of join

 Let $X$ be path-connected and $Y$ be arbitrary topological space. Then the join $X*Y$ is simply connected.

 

$\textbf{Proof}.$ We use Van Kampen style argument. 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 connected, so is $U$.

Now fix $x_0\in X$. Let $V_1$ be the cone with apex $x_0$ 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,0)\sim(x_2,y,0) $ 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.

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