Relationship among $BG, K(G,1)$ and Čech cohomology (1) If $G$ is an abelian group. We know Čech cohomology equals singular cohomology, $\check{H^1}(X,G)=H^1_{sing}(X,G)$. In addition, $\check{H^1}(X,G)$ classifies the isomorphism classes of principal $G$-bundles over $X$, does that mean the first singular cohomology group $H^1_{sing}(X,G)$ always classifies principal $G$-bundles over $X$? If it's true, it should be in the book but I never saw that..
(2) If again $G$ is an abelian group, and (1) is true, then $\check{H^1}(X,G))$=$H^1_{sing}(X,G)$= {homotopy classes of maps $X → K(G, 1)$ }, and also $\check{H^1}(X,G))$={ isomorphism classes of principal $G$-bundles over $X$ }={ homotopy classes of maps $X → BG$}. So { homotopy classes of maps $X → K(G, 1)$ }={ homotopy classes of maps $X → BG$ } as long as $G$ is abelian. But we know it is true for $G$ discrete and not in general, where am I wrong?
Thank you in advance for your help.

First, it's not true that Čech cohomology agrees with singular cohomology in complete generality. This is true for "nice" spaces. Also, I am assuming you are talking about discrete groups, otherwise I'm not sure what you mean by singular cohomology with coefficients.
Your first point is true, $H^1(X,G)=[X,K(G,1)]=[X,BG]$. You might have seen a special case of this in a textbook as something like "A line bundle is determined by its Stieffel-Whitney class".
For your second point, unless you are considering $G$ with a topology there is no contradiction, $BG$ is a $K(G,1)$.
For groups with a topology, then $BG$ is not $K(G^{\mathrm{discrete}},1)$, and so $H^1(X, G^{\mathrm{discrete}})$ doesn't necessarily have anything to do with principal $G$-bundles.
