A map to an aspherical space factorises?

Question

Let $X$ be a compact smooth manifold with fundamental group $G$ and $Y$ be an aspherical space, then why does the map $f:X\to Y$ factorise to $X\to BG\to Y$?

Answer 1

I could be wrong, but it seems that you can't choose just any $BG$ for this to be true. For example, if $X$ is simply connected, $BG$ can be just a point. If $X=Y$ is more than $0$-dimensional, then the identity map does not factor through $BG$. I will give two answers. The first is a way this works pointwise for a specific $BG$, and the other is how it works up to homotopy.

The pointwise version is that, since every smooth manifold has a CW structure, one can construct an Eilenberg-Maclane space for $X$ by starting with this CW structure and then perform the usual construction of inductively attaching $({\geq}3)$-cells to kill off any $\pi_{k\geq 2}$ that arises. This gives an inclusion $X\hookrightarrow BG$. We can inductively define a map $g:BG\to Y$ that factorizes $f$ by observing that, for some cell $e^n\subset BG$ not in $X$ whose boundary has already been accounted for, since $g(\partial e^n)$ is the image of a sphere with $n\geq 2$, it is nullhomotopic: this nullhomotopy can be used to extend $g$ into the interior of $B^n$.

The homotopy version is that, since $Y$ is aspherical, it is a $BH$ with $H=\pi_1(Y)$. The induced map $f_*:G\to H$ itself corresponds to a homotopy class of maps $BG\to BH$ (see map between classifying spaces induced by group homomorphism), and the identity map $G\to G$ corresponds to a homotopy class of maps $X\to BG$. The composition gives a homotopy class of maps $X\to BG\to Y$, which is the same homotopy class as $X\to Y$.