Please forgive me if this is not the right forum for this question.


Let $$ X = \cdots \rightarrow X\_n \rightarrow X\_{n-1} \rightarrow \cdots \rightarrow X\_0 = \ast$$ be a tower of fibrations of simplicial sets. I am trying to understand a proof of Milnor's exact sequence found in Goerss-Jardine, but I am having some trouble and have been thinking about it for quite some time. 


In the proof one considers the following model for the homotopy inverse limit of X, namely as the pullback of two morphisms where the first is $$j: \Pi\_n Hom(\Delta^1,X\_n) \rightarrow \Pi\_n X\_n \times X\_n$$ given by the inclusion $$\partial \Delta^1 \rightarrow \Delta^1$$ and the second is given by $$(1,q): \Pi\_n X\_n \rightarrow \Pi\_n X\_n \times X\_n .$$ Here , $(1,q)$ is the product of the maps $$(1,p\_n): \Pi\_n X\_n \rightarrow X\_n \times X\_{n+1} \rightarrow X\_n \times X\_n , $$ where the first map is projection followed by the identity times $p\_n$. $p\_n$ is the map $X\_{n+1} \rightarrow X\_n$ coming from the tower of fibrations. Let us call this pullback for $T(X).$ Then it is easy to see that the map $$T(X) \rightarrow \Pi\_n X\_n $$ is a fibration with fiber $$\Pi\_n \Omega X\_n .$$ Then, in the long exact sequeence in homotopy induced by this fibration, it is claimed that the boundary map $$\partial\_i: \pi\_{i+1}(\Pi\_n X\_n) \rightarrow \pi\_i (\Pi\_n \Omega X\_n) $$ can be identified, under the iso $\pi\_i(\Omega X) \cong \pi\_{i+1}(X)$' with the map $$f: \pi\_{i+1} (\Pi\_n X\_n) \rightarrow \pi\_{i+1}(\Pi\_n X\_n)$$ where $$f((a\_i))= (a\_i -p\_n(a\_{i+1})) ,$$ i.e a map whose kernel is the limit of the tower pf abelian groups $\pi\_{i+1}(X). $ 


My question is, how can the boundary map be identified with the map $f$ as above? I tried to use naturality but to no real luck. 


I would be very grateful for help.



Milnor's exact sequence and a certain proof
First take a space $X$, and try the fibration $Hom(\Delta\_1,X)\to X\times X$. The fiber over $(x,x')\in X\times X$ is the space of paths in $X$ which go from $x$ to $x'$ (which can sometimes be empty). Over a point $(x,x)$, the fiber is $\Omega X$ (based loops at $x$).


Now consider the sequence
$$
\pi\_n \Omega X \to \pi\_n Hom(\Delta\_1,X) \to \pi\_n X\times \pi\_n X \to \pi\_{n-1}\Omega X,
$$
using $(x,x)$ as a basepoint in $X$, and the constant path at $x$ in the other spaces. The middle map is really the diagonal inclusion $\pi\_n X\to \pi\_n X\times \pi\_n X$, so we get a short exact sequence
$$
0 \to \pi\_n X\to \pi\_nX\times \pi\_nX \xrightarrow{d} \pi\_{n-1}\Omega X\approx \pi\_n X \to 0.
$$ 
The only reasonable choice(s) for $d$ is $d(a,b)=ab^{-1}$ (up to reordering $a$ and $b$ and placing the $-1$) which we write as $a-b$ if the groups are abelian. Showing that this is actually true requires an actual argument, which will probably have to know exactly how the map $d$ is defined. (Or you could use naturality and the fact that $\pi\_n S^n=Z$, and the fact that these are really the only possible choices in that case.)


In the argument you describe, you are dealing with an infinite product of maps of this form as one side of your pullback square, so they work the same way.


In the homotopy inverse limit argument you describe, you actually need to use as a basepoint a path $\gamma$ from $x$ to $x'$ in $X$, which requires a little more care to read off what these groups are. ($\pi\_n (X,x)$ and $\pi\_n (X,x')$ are not the same group, but we are given a canonical isomorphism between them by the choice of $\gamma$.)


