**Proof.**
We will show that $H_ i(R(A)_\bullet ))$ and $H_ i(U(A)_\bullet )$ are canonically isomorphic to a third group. Namely

\[ \mathbf{H}_ i(A) := \frac{ \{ (a_{i, 0}, a_{i-1, 1}, \ldots , a_{0, i}) \mid d(a_{i, 0}) = \delta (a_{i-1, 1}), \ldots , d(a_{1, i-1}) = \delta (a_{0, i}) \} }{ \{ d(a_{i + 1, 0}) + \delta (a_{i, 1}), d(a_{i, 1}) + \delta (a_{i-1, 2}), \ldots , d(a_{1, i}) + \delta (a_{0, i + 1}) \} } \]

Here we use the notational convention that $a_{i, j}$ denotes an element of $A_{i, j}$. In other words, an element of $\mathbf{H}_ i$ is represented by a zig-zag, represented as follows for $i = 2$

\[ \xymatrix{ a_{2, 0} \ar@{|->}[r] & d(a_{2, 0}) = \delta (a_{1, 1}) & \\ & a_{1, 1} \ar@{|->}[u] \ar@{|->}[r] & d(a_{1, 1}) = \delta (a_{0, 2}) \\ & & a_{0, 2} \ar@{|->}[u] \\ } \]

Naturally, we divide out by “trivial” zig-zags, namely the submodule generated by elements of the form $(0, \ldots , 0, -\delta (a_{t + 1, t-i}), d(a_{t + 1, t-i}), 0, \ldots , 0)$. Note that there are canonical homomorphisms

\[ \mathbf{H}_ i(A) \to H_ i(R(A)_\bullet ), \quad (a_{i, 0}, a_{i-1, 1}, \ldots , a_{0, i}) \mapsto \text{class of image of }a_{0, i} \]

and

\[ \mathbf{H}_ i(A) \to H_ i(U(A)_\bullet ), \quad (a_{i, 0}, a_{i-1, 1}, \ldots , a_{0, i}) \mapsto \text{class of image of }a_{i, 0} \]

First we show that these maps are surjective. Suppose that $\overline{r} \in H_ i(R(A)_\bullet )$. Let $r \in R(A)_ i$ be a cocycle representing the class of $\overline{r}$. Let $a_{0, i} \in A_{0, i}$ be an element which maps to $r$. Because $\delta (r) = 0$, we see that $\delta (a_{0, i})$ is in the image of $d$. Hence there exists an element $a_{1, i-1} \in A_{1, i-1}$ such that $d(a_{1, i-1}) = \delta (a_{0, i})$. This in turn implies that $\delta (a_{1, i-1})$ is in the kernel of $d$ (because $d(\delta (a_{1, i-1})) = \delta (d(a_{1, i-1})) = \delta (\delta (a_{0, i})) = 0$. By exactness of the rows we find an element $a_{2, i-2}$ such that $d(a_{2, i-2}) = \delta (a_{1, i-1})$. And so on until a full zig-zag is found. Of course surjectivity of $\mathbf{H}_ i \to H_ i(U(A))$ is shown similarly.

To prove injectivity we argue in exactly the same way. Namely, suppose we are given a zig-zag $(a_{i, 0}, a_{i-1, 1}, \ldots , a_{0, i})$ which maps to zero in $H_ i(R(A)_\bullet )$. This means that $a_{0, i}$ maps to an element of $\mathop{\mathrm{Coker}}(A_{i, 1} \to A_{i, 0})$ which is in the image of $\delta : \mathop{\mathrm{Coker}}(A_{i + 1, 1} \to A_{i + 1, 0}) \to \mathop{\mathrm{Coker}}(A_{i, 1} \to A_{i, 0})$. In other words, $a_{0, i}$ is in the image of $\delta \oplus d : A_{0, i + 1} \oplus A_{1, i} \to A_{0, i}$. From the definition of trivial zig-zags we see that we may modify our zig-zag by a trivial one and assume that $a_{0, i} = 0$. This immediately implies that $d(a_{1, i-1}) = 0$. As the rows are exact this implies that $a_{1, i-1}$ is in the image of $d : A_{2, i-1} \to A_{1, i-1}$. Thus we may modify our zig-zag once again by a trivial zig-zag and assume that our zig-zag looks like $(a_{i, 0}, a_{i-1, 1}, \ldots , a_{2, i-2}, 0, 0)$. Continuing like this we obtain the desired injectivity.

If $\Phi : (A_{\bullet , \bullet }, d, \delta ) \to (B_{\bullet , \bullet }, d, \delta )$ is a morphism of double complexes both of which satisfy the conditions of the lemma, then we clearly obtain a commutative diagram

\[ \xymatrix{ H_ i(U(A)_\bullet ) \ar[d] & \mathbf{H}_ i(A) \ar[r] \ar[l] \ar[d] & H_ i(R(A)_\bullet ) \ar[d] \\ H_ i(U(B)_\bullet ) & \mathbf{H}_ i(B) \ar[r] \ar[l] & H_ i(R(B)_\bullet ) \\ } \]

This proves the functoriality.
$\square$

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