Let $(Y,y)$ be a CW complex (without restrictions) and $K=K(\pi,1)$ an Eilenberg-MacLane space for the abelian group $\pi$ and $*\in K$ a chosen base point. In particular, $\pi_1(K,*)\simeq\pi$ is abelian, therefore $\pi_1(K,*)\simeq H_1(K,\mathbb Z)$.
Let $[Y,K]$ be the set of classes of homotopic maps $Y\rightarrow K$ and $[Y,K]_0$ the set of classes of base homotopic maps $(Y,y)\rightarrow(K,*)$ (with $*$ any point in $K$). I want to prove that the forgetful map $$F:[Y,K]_0\rightarrow [Y,K]$$ sending a class $[f]$ into the class in $[Y,K]$ of one of its representatives, is a bijection. First we prove that $F$ is surjective and it's pretty straightforward.
Next is injectivity. Take $f,g:Y\rightarrow K$ pointed maps which are freely homotopic (so $[f]=[g]$ in $[Y,K]$). By the Homotopy axiom for homology, $f,g$ induce the same map $H_1(Y,\mathbb Z)\rightarrow H_1(K,\mathbb Z)$, hence they induce the same map $$\pi_1(Y,y)\xrightarrow{\text{Hurewicz}}H_1(Y,\mathbb Z)\rightarrow H_1(K,\mathbb Z)\xrightarrow{\simeq}\pi_1(K,*)\xrightarrow\simeq \pi$$Now, we want to use that given a $n-1$-connected CW complex $(Y,y)$, then there is an isomorphism $$[Y,K(\pi,n)]\simeq\textbf{Hom}_\textbf{Grp}(\pi_1(Y,y),\pi)\tag{*}$$But in this case we need $Y$ to be $0$-connected, which need not be the case.
This is the way I solved this problem: A CW complex is the disjoint union of its connected components (which are also its path-connected components and CW subcomplexes), so restricting $f,g$ to the path component $C$ of $y$, we still have $f|_{C}$ homotopic to $g|_C$, hence we can simply repeat the argument and apply (*) to reach the conclusion $f|_C$ is base homotopic to $g|_C$, then extend this base homotopy $H:C\times I\rightarrow K$ to an homotopy $J:Y\times I\rightarrow K$ by taking $J=H$ in $C\times I$ and $J$ equal to the free homotopy between $f, g$ in all the other components.
Is this argument correct? I'm pretty sure it is, but maybe I missed something.
