4

Let $X,Y$ be CW-complexes with $X$ finite dimensional and $X = \bigcup_{n \in \Bbb N} X_n$ where the $X_n\subset X_{n+1}$ are finite sub-complexes of $X$. If $f: X \rightarrow Y$, with $f|_{X_n}$ null-homotopic is $f$ necessarily null-homotopic?

EDIT: Please do not downvote existing answers as I had forgotten to include the assumption $X_n\subseteq X_{n+1}$

PVAL-inactive
  • 8,799

3 Answers3

4

Suppose $X=Y=S^1$ with its CW-decomposition with two vertices and two $1$-cells, $f:X\to Y$ the identity map, and let the $X_n$ be all its proper subcomplexes.

Mariano Suárez-Álvarez
  • 139,300
  • Notice why this works: every one of my $X_n$ is contractible. You can do this with any noncontractible $X$ by taking the $X_n$ to be its cells... – Mariano Suárez-Álvarez Jan 30 '15 at 20:59
  • I suspect the intent was that $f$ be null-homotopic for every finite subcomplex $X_n$. –  Jan 30 '15 at 21:03
  • The cases I'm interested in are all when $X$ to be a direct limit of $X_n$'s. I thought what I wrote was equivalent, but this answer shows I was off. Do you know what happens under the additional assumption $X_n\subset X_{n+1}$? – PVAL-inactive Jan 31 '15 at 01:37
  • Notice that my example is the direct limit of its proper subcomplexes (the limit is not directed though, which is what you had in mind probably) If it is an increasing chain, I'd imagine it does work. – Mariano Suárez-Álvarez Jan 31 '15 at 01:46
  • Really the motivation came out of the last part of this attempted answer: http://math.stackexchange.com/questions/1107682/elementary-proof-of-the-fact-that-any-orientable-3-manifold-is-parallelizable/1108061#1108061. The idea was if we have some classifying map for say open (some structure) manifolds, and we know that this classifying map is trivial on all compact (some structure) manifolds with boundary and we can build the first out of the latter as a filtration the whole map is trivial. – PVAL-inactive Jan 31 '15 at 01:59
4

There are examples of fantom maps from finite-dimensional CW-complexes, see for instance

C. McGibbon, Phantom maps. In: I. James (ed), "Handbook of algebraic topology", pp. 1209-1257.

The examples themselves appear on page 1212. The space $X$ is the mapping telescope of a sequence of maps $S^n\to S^n$ whose degrees are coprime to a fixed prime $p$. McGibbon refers to these examples as "fantom maps of the 2nd kind". For the purpose of the question, these are maps which are not null-homotopic, whose restrictions to each finite subcomplex is null-homotopic.

Edit. Here is a useful theorem:

For a CW complex $X$ and a number $n$ we define the subgroup $Fantom^n(X)< H^n(X)$ to consist of (fantom) cohomology classes whose restriction to each finite subcomplex in $X$ is trivial.

Theorem. $Fantom^n(X)\cong Ext^1(H_{n-1}(X)/Torsion, {\mathbb Z})$.

See Infinite CW-complexes, Brauer groups and phantom cohomology, page 2.

Thus, one can construct a space with nonzero 2nd fantom cohomology by taking, for instance, a space with $H_1(X)\cong {\mathbb Q})$ (since $Ext^1({\mathbb Q}, {\mathbb Z})$ is enormous, isomorphic to ${\mathbb A}/{\mathbb Q}$, where ${\mathbb A}$ is the group of adeles). The space $X$ can be taken 2-dimensional, a presentation complex for the additive group of rational numbers.

To get from fantom cohomology to fantom maps, look at $[X, K({\mathbb Z}, 2)]$.

Moishe Kohan
  • 111,854
2

If instead of demanding that $X$ is finite-dimensional, you demand that $Y$ is finite-dimensional, a counterexample is constructed in Adams, Walker, "An example in homotopy theory." An $f: X \to Y$ is constructed such that $f\big|_{X^n}$, the restriction to the $n$-skeleton, is null-homotopic, but $f$ itself is not. (The $X$ here is $\Sigma \mathbb{CP}^\infty$, and the $Y$ is homotopy equivalent to a countable wedge of 4-spheres.) Because every finite subcomplex is contained in some $n$-skeleton, this shows that $f$ is null-homotopic when restricted to any finite subcomplex of $X$.

  • 1
    The famous phantom map. – Najib Idrissi Jan 30 '15 at 21:08
  • I was aware of this, in the examples I care about $Y$ is some classifying space and $X$ is a manifold and $f$ is the classifying map. The reason to demand $X$ f.d. was to avoid these examples (though I do not know if that doesnt mean there arent others). – PVAL-inactive Jan 31 '15 at 02:01