Role of $0$ in Hatcher's proof of higher homotopy groups distributivity In Hatcher 4.1, page 341 in my edition Hatcher defines the following. Let $f \in \pi_n(X,x_0)$ and a path $\gamma$ from $x_0$ to $x_1$. Then, $\gamma f$ if defined "by shrinking the domain of $f$ to a smaller concentric cube in $I^n$, then inserting the path $\gamma$ on each radial segment in the shell between this smaller cube and $\partial I^n$"
Hatcher then proves that $\gamma( f + g) = \gamma f + \gamma g$. He writes:

For $(1)$, we first deform $f$ and $g$ to be constant on the right and left halves of $I^n$, respectively, producing maps we may call $f+0$ and $0+g$, then we excise a progressively wider symmetric middle slab of $\gamma(f+0) + \gamma(0+g)$ until it becomes $\gamma(f+g)$.  

He has a nice picture of this too. I think I get basically what is going on here, but I must be wrong because I do not see why we need to "add zero" to both maps. I do not understand what the role is is of the constant slap on the right hand of $f$ and the left hand of $g$. Why can we not just put the cubes together and "excise the middle slab" without having the constant map piece in the middle?
If my question does not make sense, please let me know.
 A: As far as I can tell, this is not actually necessary in any way, and is just done for convenience so that there is a clean explicit formula for the homomtopy from $\gamma(f+0)+\gamma(0+g)$ to $\gamma(f+g)$, namely:
$$h_t(s_1,s_2,\dots,s_n)=
\begin{cases}
\gamma(f+0)((2-t)s_1,s_2,\dots,s_n)) & s_1\in[0,1/2] \\
\gamma(0+g)((2-t)s_1+t-1,s_2,\dots,s_n) & s_1\in[1/2,1]
\end{cases}$$
In particular, note that Hatcher's formula neatly avoids having to get into messy details about the precise parameterization of $\gamma f$.  It relies on the fact that for any $t\in[0,1]$, the two cases in the definition of $h_t$ agree at $s_1=1/2$, since the first case comes from the right half of $f+0$ and the second case comes from the left half of $0+g$.
If you didn't replace $f$ with $f+0$ and $g$ with $g+0$, then instead of just homotoping away the right half of $\gamma(f+0)$ and the left half of $\gamma(0+g)$, you would have to homotopy away the right part of $\gamma f$ and the left part of $\gamma g$ where the path $\gamma$ is being used, and worry about doing so in such a way that the parameterization of the outer $\gamma$ shell at the end of the homotopy is exactly the same as the one used in $\gamma(f+g)$.  Notice in particular that if you look at a point in the right half of the top edge of the little $f$ cube inside $\gamma f$, its $\gamma$ path to the boundary tilts to the right.  On the other hand, after homotoping to $\gamma(f+g)$, this point will be in the left half of the little cube, and so its $\gamma$ path will need to tilt to the left.  So your homotopy would need to be carefully set up to adjust all these $\gamma$ paths appropriately.  This can be done, but it's a bit messy, and requires actually writing an explicit formula for $\gamma f$ which Hatcher is avoiding.
