Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

\begin{equation} \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} \end{equation}

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

Here is a reference for whitehead towers:


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


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!

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.