Lemma 11.42 Rotman's algebraic topology 
This is lemma 11.42, pg 355, of Rotman's Algebraic Topology. 
  The context is we are trying to determine an explicit map for "connecting homomorphism". 



where $i:A \rightarrow X$ is inclusion. This is Rotman's proof. 



Then there is calculations here giving explicit map $\theta:[S^n, Mi] \rightarrow \pi_{n+1}(X,A)$ 




The last statement is unclear to me. He claims we have all the maps in the original diagram. 
But between $\pi_{n+1}(X)$ and $[S^n,\Omega X]$ we have the series of isomoprhisms
  $$\pi_{n+1}(X)=[S^{n+1},X] \cong [\Sigma S^n, X ] \cong [S^n, \Omega X]$$
  where the first, albeit natural, isomorphism, has a choice. 
Hence, the question is how is the map $(\Omega^nk)_*$ actually defined pointwise? Its domain should be $[S^{n+1}, X]$ instead of $[S^n, \Omega X]$. 


Here is our map: 
$$\Omega^nk_*:[S^{n+1}, X] \xrightarrow{\varphi_*} [ \Sigma S^n, X] \simeq [ S^n , \Omega X] \xrightarrow{k_*} [S^n, Mi]$$ 
The first map clearly depends on our choice $\varphi$. At which point is $\varphi_*$ cancelled again? 

All right, I made a mess a bit in the comments. Let me make it clear with diagrams.
Note that by iterating as many as you want, we can just consider the adjunction $[S^{n+1},X]\to [S^n,\Omega X]$ instead of iterated adjunction as in Rotman. Actually, this does not have exactly the form of adjunction; instead it is a composition of $(\phi^{-1})_*:[S^{n+1},X]\to [\Sigma S^{n},X]$ and the usual adjunction $[\Sigma S^n,X]\to [S^n,\Omega X]$ where $\phi:S^{n+1}\to \Sigma S^n$ is your favorite identification between $S^{n+1}$ and $\Sigma S^n$. And there is no particular choices made in horizontal arrows. 
But note that the two-row diagram is in fact a three-row diagram:

(I just realized that I reversed all the vertical arrows, my bad. It doesn't affect underlying math as vertical parts are isomorphisms though.)
Here the horizontal arrows are induced by continuous maps acting on the second entry of the bracket. Here the commutativity of the first square is standard. (At any rate, it is the standard adjunction.) And the commutativity of the second square is just obvious; it just says the bifunctoriality of the bracket $[-,-]$.
And of course the dependency on $\phi$ is only on the second square. 
