Suppose we have a map $f:S^1 \wedge S^1 \to X$ of pointed spaces ($S^1$ is the circle), let $T:S^1 \wedge S^1$ be the map that flips the factors, (so $T(x,y)=(y,x)$) and let $f'=f \circ T:S^1 \wedge S^1 \to X$. My question is why is $[f + f' ]=0$ in the homotopy group $\pi_2(X) =[S^1 \wedge S^1, X]$?
I believe this is true because often times when dealing with suspension and spectra, there are minus signs introduced when flipping the $S^1$ factors, and I believe the above reason is the fundamental result why we need to do so.
Unwinding the definition of $+$ in $\pi_2(X)$, here is what my question boils down to: Let $I=[0,1]$ and think of $S^1 \wedge S^1$ as $I^2/\partial I^2$, and let $f:I^2 \to X$ be a map sending $\partial I^2$ to the base-point $x_0 \in X$ (and so it induces a map $[f]:S^1 \wedge S^1 \to X$). Define a map $f+f':I^2 \to X$ by $$ (f+f')(s,t)= \left\{ \begin{array}{ll} f(2s,t) & s \in[0,1/2] \\ f(t,2s-1) & s\in [1/2,1], \text{note } f'(2s-1,t)=f(t,2s-1) \end{array} \right. $$
Note that $f+f'$ sends $\partial I^2$ to $x_0$, so induces a basepoint-preserving map $[f+f']:S^1 \wedge S^1 \to X$.
I want to show $[f+f']$ is nullhomotopic, i.e I need to construct a homotopy between $f+f'$ and the constant map $c:I^2 \to X$ whose image is $x_0 \in X$, i.e a function $H(s,t,u):I^2 \times I \to X$ such that $H(s, t, 0)=f+f'$, $H(s,t, 1)=c$, and $H|(\partial I^2 \times I)$ is the constant map with image $x_0$ (I was having trouble getting the last part to hold under any homotopy I was making up).