# A counter example in obstruction theory

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.

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This is most probably a silly comment. I´m not in obstruction theory but I would like to understand the problem. When you ask for a map $f: K^n \rightarrow Y$ you want it to be continuous, don´t you? Then the first example which came into my mind was $K$ to be the "full" tetraedron (i.e. the complete simplicial complex on 4 vertexes), $Y$ to be $K^2$ as a topological space and finally $f^1: K^1 \rightarrow Y$ to be the inclusion map. Unfortunately I don´t understand why it does not work as a conuterexample. I would be glad to know where does it fail... – Giovanni De Gaetano Apr 2 '12 at 13:53
@GiovanniDeGaetano, you understand correctly, a map in topology is always assumed to be a continuous function. The example you thought of fails because there are different ways to extend your $f^1$ to $f^2$. The obvious one, namely the identity, does not have an extension to $f^3$, but there is a different $f^2$, a null-homotopic one, that can be extended to $f^3$. In general, if $f^1$ is null-homotopic as in your case, it can always be extended to $K$ by the homotopy extension property for simplicial pairs. – KotelKanim Apr 2 '12 at 19:58
I´m sorry if I waste your time with dumb questions. But I don´t see why $f^1$ is null-homotopic. Indeed if it was then also the map $f^1: K^1 \rightarrow Image(f^1)=K^1$ (which is $id_{K^1}$) would be null-homotopic and then the space $K^1$ would be contractible. But it is not. In specific I don´t see the null-homotopic extension of $f^1$, may you describe it? Thank you! – Giovanni De Gaetano Apr 3 '12 at 8:32
No problem. $f^1$ is null-homotopic because $Y$ is homeomorphic to a sphere and since $f^1$ is not onto, its image contained in the sphere minus a point which is homeomorphic to a disc and thus contractible. Any map to a contractible space is null-homotopic (just compose with the "contraction"). The conclusion that this must imply that $K^1$ is contractible is wrong since a subspace of a contractible space does not have to be contractible (think of the inclusion of the sphere into the euclidean space). – KotelKanim Apr 3 '12 at 8:52
@Qiaochu This was x-posted to MO and answered there ($X=\mathbb RP^3$, $Y=\mathbb RP^2$) – Grigory M Jan 3 '15 at 10:09

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 $x^3 \neq 0$.