Let $X, Y, Z$ be based spaces and define
$F_2(X, Y, Z) = \{(x, y, z) \in X \times Y \times Z|$ at least one of $x, y, z$ is $= * \}$
Prove that the inclusion $F_2 (X, Y, Z) \to X \times Y \times Z$ is surjective on homotopy groups.
Using that long exact sequence for homotopy groups I see that this is the same as showing that $j_* : \pi_{n}(X \times Y \times Z, *) \to \pi_{n}(X \times Y \times Z, F_2(X, Y, Z), *) $ is the zero map where $j$ is the natural inclusion as it appears in the long exact sequence for homotopy groups.
Now this means that any map from $I^{n} \to X \times Y \times Z$ which maps the boundary to the base point can be homotoped fixing the base point to a map whose image lies in $F_2(X, Y, Z)$. This makes intuitive sense as the image of the $I_n$ can be stretched to the 'walls of the cube' (cube here is $X \times Y \times Z$). The holes are not a problem because they arise from holes in either of the spaces and hence extend to cylinders in the cube' $X \times Y \times Z$ and fixing the image of the boundary is not a problem since it is just the origin.
Please help me.