8
$\begingroup$

Let $K$ denote a simplicial complex and $Y$ some topological space. Let us also denote by $K^n$ the $n$-skeleton of $K$. I would like to have an example for the following situation:

There is a map $f^1:K^1\to Y$ that can be extended to $f^2:K^2\to Y$ and yet no such extension can be further extended to $f^3:K^3\to Y$.

The idea is that there is an obstruction to the existence of $f^3$ already on the one-dimensional level but not by obstructing the existence of $f^2$. It is written in Hilton and Wylie's book that it is possible, yet I was not able to construct an explicit example myself.

  • 5
    @Qiaochu This was [x-posted to MO](http://mathoverflow.net/q/93098/) and answered there ($X=\mathbb RP^3$, $Y=\mathbb RP^2$)2015-01-03

1 Answers 1

2

This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Neil Strickland below.

Here is an example with CW complexes rather than simplicial complexes. I doubt that there is an important difference, although the simplicial case will require more bookkeeping.

Take $K = \mathbb{R}P^3$ and $Y = \mathbb{R}P^2$. We can give $K$ a CW structure with skeleta $\mathbb{R}P^k$ for $0 \leq k \leq 3$. Let $f^1 : \mathbb{R}P^1 \to Y$ be the evident inclusion. Clearly this extends over $K^2$. Now suppose we have an extension $f^3 : K^3 = K \to Y$ of $f^1$. This will then give a graded ring homomorphism $(f^3)^* : H^*(Y ; \mathbb{Z}/2) \to H^*(K; \mathbb{Z}/2)$, or in other words $(f^3)^* : (\mathbb{Z}/2)[y]/y^3 \to (\mathbb{Z}/2)[x]/x^4$. Because $f^3$ extends $f^1$ we must have $(f^3)^*(y) = x$. This gives a contradiction because $y^3 = 0$ but $x^3 \neq 0$.