$H_{n+1}(K(Z_m,1),Z) \to H_{n+1}(K(Z_m,1),Z/m)$ is surjective for $n$ odd In hatcher, on the first example(ex. 3E.1) on the page for bockstein homomorphisms, he uses the fact that the reduction map $H^{n+1}(K(Z_m,1),Z) \to H^{n+1}(K(Z_m,1),Z/m)$ is surjective.  I know this is true because(W.L.O.G lets do this for $n=1$), the cellular cochain groups with $Z$ coefficients look like $Z  \xrightarrow{0} Z \xrightarrow{m} Z \xrightarrow{0} Z...$ where I put the cochain groups in ascending order with respect to the grading.
Mod $m$ these are all $0$ maps.  Now the reduction map is surjective if a mod $m$ cocycle is still a cocycle for some lift to a cocycle with $Z$ coefficients.  This happens here because map from $C^2(K(Z_m,1),-)$ to $C^3(K(Z_m,1),-)$ is the zero map for coefficients in $Z$ and $Z_m$.
Now I am sure this is not the way that Hatcher wanted me to see it because he avoids using the cell structure for everything other than computing the cohomology.  So is there a way to do this using universal coefficient theorem?
 A: What you want to know follows just from knowing the cohomology groups and the long exact you get from $\mathbb Z \to \mathbb Z \to \mathbb Z/m$.
The map $\mathbb Z \overset{m}\to \mathbb Z$ of multiplication by $m$ always induces multiplication by $m$ on $H^i(X; \mathbb Z)$. Since $H^i(X; \mathbb Z) = \mathbb Z/m$ or $0$, in this case it is the zero map. So the long exact sequence breaks into short exact sequences$$0\to H^{n+1}(X; \mathbb Z) \to H^{n+1}(X; \mathbb Z/m) \to H^{n+2}(X; \mathbb Z) \to 0,$$ but only one of $H^{n+1}(X;\mathbb Z)$ and $H^{n+2}(X;\mathbb Z)$ is non-zero, so the desired map is an isomorphism or zero.
A: Let $X=K(Z_m,1)$.  This is a twist on Ben's answer.  Take the long exact sequence of the Bockstein associated to $0 \to \mathbb{Z} \to \mathbb{Z} \to \mathbb{Z}/m \to 0$.  Note that $H^{n+2}(X,\mathbb{Z})=0$ so the connecting homomorphism $\tilde \beta=0$ coming from $H^{n+1}(X,\mathbb{Z}/m)$ is 0 and the reduction map is surjective. 
The reason why the existence of a $Z$ cocycle for any mod-m cocycle implies that 
This is essentially the same argument that I was using with the cell structures:  the reduction map is surjective, is that the boundary map $\tilde \beta=0$ coming from $H^{n+1}(X,\mathbb{Z}/m)$ is 0.  The boundary map will be zero under this condition because the boundary map is defined by taking a representative cochain, taking the coboundary and dividing by $m$, so if the representative cochain is already a cocycle, then when you take the coboundary, it will be zero.
