I'm looking for $\pi_1(S(K\lor T))$; I know that for klein bottle $K$ and torus $T$ we can represent each of these as as
$$K = S^2 \lor S^1 \lor S^1$$ and
$$T = S^1 \times S^1, $$ which hasn't helped, and I know the representation of the fundamental group of $K$ as $\langle a,b \mid aba^{-1}b\rangle$ and the fundamental group of the torus is $Z \times Z$, but I really just am at a total loss of what to do. I've tried looking at them as cell complexes, but once I wedge them I have no idea how to go about suspending the connected sum. Been thinking about this problem for a solid month and I finally have some time, so any help is much appreciated.
Best Answer
The two spaces $K$ and $T$ are both path connected and so does $K\vee T$, since any two point $a\in K, b\in T$ and be connected by a composite path $f*g$ from $a$ to $b$ where $f(0)=a,f(1)=g(0)=x_0,g(1)=b$ ($x_0$ is the basepoint and also the common point).
Denote $K\vee T$ by $X$.
The suspension of X can be obtained by a quotient map $q:X\times I\to SX$. Now the basepoint $x_0$ is sent to $x'_0=q(x_0,\frac{1}{2})$. Then Consider two open path connected subspace of $I$ ,they are $(m,1]$ and $[0,n)$ where $m\in(0,1/2),n\in(1/2,1)$, then $ x'_0\in q(X\times (m,1])\cap q(X\times [0,n))$.
Now, let $A=q(X\times (m,1])$ and $B=q(X\times [0,n))$, both of them are contractible because we can slide each point through a path pointing to $SX\times\{1\}$ and $SX\times \{0\}$, respectively. Take $A$ as an example, it can be contract by $$ G_A((x,s),t)=(x,(1-s)t+s) $$ A similar construction works for $B$. So, $\pi_1(A,x'_0)\approx\pi_1(B,x'_0)=0$ and by Seifert-Van Kampen Thm (we can use it because $X$ is path-connected), $ \pi_1(SX,x'_0)$ is trivial.