Whitehead product $[\alpha, \beta]=0$ iff there exists an extension. I am having a lot of difficulty understanding the Whitehead product so I think proving an "obvious" result would help with my understanding.
However, I am still completely at a loss as how to show that $[\alpha, \beta]=0$ if and only if there exists a map $F: S^k \times S^l \to X$ such that $F \circ i_1 \cong \alpha$ and $F \circ i_2 \cong \beta$. 
I think that $\alpha \vee \beta$ extending to a map $F: S^k \times S^l \to X$ implies that $\alpha \vee \beta$ extends to a map on $D^{k+l}$ in which case it would need to be trivial. 
I would really appreciate the help as I am really struggling to work with the Whitehead product. 
 A: Let's fix nome notations : I write $i = i_1\vee i_2: S^k\vee S^l\to S^k\times S^l$, $j:S^{k+l-1}\to D^{k+l}$ the inclusion map, and $\varphi: S^{k+l-1}\to S^k\vee S^l$ the canonical attachment map. 
By definition, $S^k\times S^l$ is obtained by attaching $D^{k+l}$ to $S^k\vee S^l$ along $\varphi$, so by definition we have the universal property : for any space $Y$, if $f: D^{k+l}\to Y$ and $g: S^k\vee S^l\to Y$ satisfy $g\circ \varphi = f\circ j$, then there is a unique $h: S^k\times S^l\to Y$ such that $f = h\circ p$ and $g = h\circ i$, where $p: D^{k+l}\to S^k\times S^l$ is the map resulting from the attachment.
Recall that if $\alpha: S^k\to X$ and $\beta: S^l\to X$, then by definition $[\alpha,\beta]$ is the homotopy class of $(\alpha\vee \beta)\circ \varphi: S^{k+l-1}\to X$.
Suppose there is such an $F$. Then this means that $\alpha\vee \beta = F\circ i$, so $(\alpha\vee \beta)\circ \varphi = F\circ i\circ \varphi$. But now $i\circ \varphi: S^{k+l-1}\to S^k\times S^l$ is homotopically trivial because $i\circ \varphi = p\circ j$ and $D^{k+l}$ is contractible, so $[\alpha,\beta]=0$.
Conversely, if $[\alpha,\beta]=0$, then it means that $(\alpha\vee \beta)\circ \varphi: S^{k+l-1}\to X$ is homotopically trivial, so it extends to a map $f: D^{k+l}\to X$. Now we can apply the universal property above with $Y=X$ and $g = \alpha\vee \beta$ to get some $F: S^k\times S^l\to X$ satisfying the factorisation we wanted.
