# extending maps from spaces to their whitehead towers

Let $f \,: X \to Y$ be a map between connected spaces. Let: $$X^{(k)} \to \ldots \to X^{(0)} \approx X$$ and $$Y^{(k)} \to \ldots \to Y^{(0)} \approx Y$$ be whitehead towers for $X$ and $Y$. What are sufficient conditions in order to extend $f$ to a "morphism" $\{f_{n}\}_{n \in \mathbb{N}}$ between the whitehead towers? i.e., in such a way that the following diagram commutes?

$$\label{} \begin{array}{ccccccccccccccccccccccccccccccc} \vdots & & \vdots \\ \Big\downarrow && \Big\downarrow \\ X^{(2)} & \overset{f_2}{\longrightarrow} & Y^{(2)} \\ \Big\downarrow && \Big\downarrow \\ X^{(1)} & \overset{f_1}{\longrightarrow} & Y^{(1)} \\ \Big\downarrow && \Big\downarrow \\ X & \overset{f}{\longrightarrow} & Y \\ \end{array}$$

Maybe stronger hypotheses are needed: the maps of the whitehead towers may be fibrations for example (or pincipal bundles.. whatever!)!

EDIT

maybe this question was already answered there: http://mathoverflow.net/questions/5268/functorial-whitehead-tower ... but maybe by "functorial" they simply mean "canonical", i.e., it is possible to associate to a space a distinguished whitehead tower!

EDIT

it seems to me that proposition 4.18 p. 364 of hatcher's "algebraic topology" provides an affirmative answer to my question (if the vertical maps in the towers are fibrations). Translateing the notation of the book to my case, we get $A = \emptyset$, $Z = X^{(n+1)}$, $X = X^{(n)}$, $Z' = Y^{(n+1)}$, $X' = Y^{(n)}$, $f$ and $f'$ are fibrations, and $g = f_{n-1}$. Hence there exists a map $h \,: Z \to Z'$ as in the proposition, because $X^{(n+1)} \to X^{(n)}$ is a $n + 1$--connected model for $X^{(n)}$ (see hatcher's definition: p. 352)!

The diagram of proposition 4.18 is then commutative only up to homotopy. But since $f$ and $f'$ are fibrations (in our case), we are able to "correct" $h$ alias $f_{n+1}$, making the diagram genuinely commutative!

-