Double complex and UCT I am new to study spectral sequences and double complexes, but this question arose while I was trying to do some computation for a project.   
Imagine you have a double complex $\{C_{ij}, d_1,d_2\}$ and that the spectral sequence on it converges to the homology of the total complex and where all the $C_{ij}$'s are free abelian groups. Now, let $R$ be a PID and define $C^*_{ij}:= \text{Hom}(C_{ij};R)$ and $d^1$ and $d^2$ are the dual of the original map (basically I dualise the groups and reverse all the arrows). Then, assume also that the spectral sequence of $\{C_{ij}^*, d^1,d^2\}$ converges to the homology of its total complex. Are these groups corresponding to the cohomology groups of the original total complex? And if so, could I have found them by applying the Universal Coefficient Theorem for cohomology?  
Thanks in advance!
 A: So first of all let me note that this has very little to do with spectral sequences : you have a double complex $(C_{ij})$, on the one hand you take $Tot(C_{*,*})$ and then its cohomology with coefficients in $R$, and on the other hand you first take $C^{*,*}:= \hom(C_{*,*}, R)$, then $Tot(C^{*,*})$ and then the homology of that, and ask whether the two things coincide. 
The answer is that they won't in general, but they will in a lot of practical cases (e.g. if $C_{*,*}$ is first quadrant - you'll see clearly from the explanation what the real condition is)
Indeed if you look at the first construction, it looks like this : start with $C_{p,q}$, build $K_n = \bigoplus_{p+q=n}C_{p,q}$, then $K^n := \hom(K_n, R)\cong \prod_{p+q=n}\hom(C_{p,q},R) \cong \prod_{p+q=n}C^{p,q}$, and you take the homology of that.
The other construction on the other hand starts with $C_{p,q}$, builds $C^{p,q}:=\hom(C_{p,q},R)$ and then $H^n = \bigoplus_{p+q=n}C^{p,q}$. 
So $K^n$ and $H^n$ look a lot like one another, and in fact

If for all $n$, $\{(p,q)\mid p+q=n \land C^{p,q}\neq 0\}$ is finite, $K^*\cong H^*$, and therefore the two homologies coincide. 

This just follows from the fact that a finite direct sum is the same as a finite direct product. This doesn't hold in general if the indexing set is infinite though, and so 

In general, the two homologies do not coincide. 

I will provide an example for this second scenario. Before doing so, let me note that if $C^{p,q}=0$ if $p<0$ or $q<0$, then the set in question is obviously finite, and therefore the two homologies agree. Therefore :

If $(C_{*,*})$ is a first quadrant double complex, the two homologies agree. 

Note that this is clearly not an "if and only if" statement. 
Now for the example :
Say $C_{p,q} = D_p$ for some chain complex $D_*$, with vertical differential $=0$. 
Then $K^n = \prod_{p+q=n}\hom(D_p,R) = \prod_{p\in \mathbb Z}\hom(D_p,R)$ and $H^n = \bigoplus_{p\in \mathbb Z}\hom(D_p,R)$, and the differential is, in both cases, the obvious one. 
Then (since products and direct sums are exact in $R-\mathbf{Mod}$) $H_n(K^*) = \prod_{p\in\mathbb Z}H_p(D)$, and $H_n(H^*) = \bigoplus_{p\in\mathbb Z}H_p(D)$
Clearly these are different for a general $D$. 
