Is it always true that a continuous map $f : X^n \to Y$ extends to $X^{n+1} \to Y$?

Question

Let $X$ be a CW complex and let $X^i$ denote the $i^{\text{th}}$ skeleton of the CW complex. Is it always true that a continuous map $f : X^n \to Y$ extends to $X^{n+1} \to Y$? In my view, since the attaching map $\phi_{\beta}$ of any $n+1$ cell $e^{n+1}_{\beta}$ is always null-homotopic via the characteristic map of $e^{n+1}_{\beta}$, $f_*([\phi_{\beta}])$=0, hence we can extend $f$ over the cell $e^{n+1}_{\beta}$. Is that right?

Answer 1

Let $X$ be the closed disc and $Y = S^1$; note that $X^1 = S^1$. Consider a continuous map $f : S^1 \to S^1$. This map extends to the disc if and only if $f$ is nullhomotopic. So for example, $f = \operatorname{id}_{S^1}$ doesn't extend to the disc.

Answer 2

Take $X$ to be a three ball with a CW complex structure having one vertex, one 2-cell and one 3-cell. Take $Y$ to be the 2-skeleton of $X$. Then you cannot extend the identity $f\colon X^{(2)}\to Y$ to a continuous map on $X$.

Homology theory is well suited to give conditions on when such an extension is possible. For example is $Y$ is contractible, then you can always extend such a map.