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Lemma: Suppose we have a $n$-dimensional CW complex $X$ and an aspherical space $Y$, where aspherical just means $\pi_n(Y)=0$ for all $n\geq 0$. Let $f,g:X\rightarrow Y$ be continuous maps. Then $f$ and $g$ are homotopic.

So my question is how to prove this assertion. Now to simplify the proof it suffices to show that $f$ is nullhomotopic. Now I can see why this is true since if we look at the attachment maps on $X$ and compose them with $f$, they are nullhomotopic from our aspherical assumption, so we can make the $n$-cells constant throughout a homotopy, but I don't know how to formally show this fact since the $n-1$ skeleton would have to be perturbed slightly to form this homotopy. Can anyone clarify what I am trying to do in a more formal setting.

Enigma
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1 Answers1

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Basically you want to extend your map $f:X\to Y$ to a map $F\colon X\times I\to Y$, where the restriction to $t=0$ is $f$ and the restriction to $t=1$ maps to a point $p\in Y$. You do this by building up the map on $k$-skeleta $X^{(k)}$. First you look at the $0$-skeleton crossed with $I$, $X^{(0)}\times I$. $X^{(0)}\times\{0\}$ maps via $f$ to some points in $Y$. Connect these by paths in $Y$ to $p$. This gives you $F|X^{(0)}\times I$. Now we need to extend $F$ to $2$-cells in the product. The boundaries of the $2$-cells are homeomorphic to circles and map into $(X\times I)^{(1)}$ and then into $Y$. These are null-homotopic in $Y$, so we can extend $F$ across the $2$-cells. Similarly we extend to $3$-cells, since their attaching maps are maps of $2$-spheres which, when mapped into $Y$ are null homotopic. Proceed inductively.

Cheerful Parsnip
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