I want to prove that a Postnikov tower is unique up to homotopy equivalence.
Edit Let us also assume all spaces to be $CW$-complexes.
A Postnikov tower is defined (on a base space $X$) as maps $X \rightarrow X_n$ and connecting maps $X_n \rightarrow X_{n-1}$ for which the following hold:
- The map $X \rightarrow X_n$ induces an isomorphism on $\pi_i$ for all $i < n$
- $\pi_i(X_n) = 0$ for all $i > n$
Suppose there exists two such towers $X_1, \cdots , X_n$ and $\overline{X_1}, \cdots , \overline{X_n}$.
There is a commutative diagram as follows: $$\require{AMScd} \begin{CD} X @>f>> X_n\\ @. {_{\rlap{h}}\style{display: inline-block; transform: rotate(30deg)}{{\xrightarrow[\rule{4em}{0em}]{}}}} @VVgV\\ @. \overline{X_n} \end{CD}$$
Passing the $\pi_i$ functor we get isomorphisms for each of the arrows (in other words the map $\pi_i(g):\pi_i(X_n) \rightarrow \pi_i(\overline{X_n})$ is well defined and indeed an isomorphism) This makes all the towers associated with a base space $X$ to be equivalent up to homotopy.
I believe that this proof is flawed. Is there any way I can use an analogous argument?
Edit: Let us define the map $g$ such that it proves weak homotopy equivalence and then maybe I can use the Whitehead's theorem which says that for $CW$- complexes homotopy equivalence is the same as weak homotopy equivalence.
Let $f_*: \pi_i(X) \rightarrow \pi_i(X_n)$ and $h_*:\pi_i(X) \rightarrow \pi_i(\overline{X_n})$. Now if $\require{AMScd} \pi_i(X) \ni [\alpha] \mapsto [\beta] \in \pi_i(X_n)$ (under $f_*$) and $\require{AMScd} \pi_i(X) \ni [\alpha] \mapsto [\sigma] \in \pi_i(\overline{X_n})$ (under $h_*$) then $\require{AMScd} \pi_i(X_n) \ni [\beta] \mapsto [\sigma] \in \pi_i(\overline{X_n})$ (under $g_*$)
Thus making it weakly homotopy equivalent?
Edit: To see that $g_n$ indeed exists with the following properties we shall define it to be $h_n(f_n^{-1})$. Since the homotopy functor $\pi_i$ respects compositions (and $\pi_i(f)$ and $\pi_i(h)$ are both isomorphisms) we have that $\pi_i(g_n)$ is also an isomorphism.
I don't know how to frame it in a better way but here is an example: $$G \overset{\cong} \longrightarrow \overline{G} \overset{\cong} \longrightarrow K$$ where both the morphisms are isomorphisms. Now since $\overline{G} \rightarrow K$ is an isomorphism so is its inverse $K \rightarrow \overline{G}$ and similarly for the pair $G$ and $\overline{G}$ we see that the composition $G \rightarrow K$ should also be an isomorphism.
However I believe I am assuming some things (like continuity of $f,h$ etc). Is there any way to wind things up?