Undertstanding Cup products I am trying to understand cup products from the following notes and am having some difficulty understanding theorem which in a way defines cup product for cohomology: https://www.math.ucla.edu/~sharifi/groupcoh.pdf#theorem.1.9.5
What does it mean to say that the map 'cup products'
$$H^i(G, A)\otimes _{\mathbb Z} H^j(G,B) \xrightarrow{\smile} H^{i+j}(G, A \otimes _{\mathbb Z}B)   $$
is "natural on $A$ and $B$" ?
And how do we get the commutative diagram on the top of page 37  ( https://www.math.ucla.edu/~sharifi/groupcoh.pdf#page.37 ) ?
I believe it has something to do with the 'naturality'.
Any help is appreciated and feel free to give any reference.

It means naturality in the sense of category theory.  Saying the cup product is natural in $A$ means that for any $B$ and any map $f: A \to A'$ you get induced maps $H^i(G, A)\otimes_{\mathbb{Z}} H^j(G, B) \to H^i(G, A')\otimes_{\mathbb{Z}} H^j(G, B)$ and $H^{i+j}(G, A\otimes_\mathbb{Z} B) \to H^{i+j}(G, A' \otimes_{\mathbb{Z}} B)$, and the square formed by these maps and the cup product commutes.
$\delta$ here is the connecting homomorphism arising from whichever short exact sequence is under consideration. The vertical maps in the diagram at the top of p.37 are $\delta \otimes 1$ on the left, and $\delta$ - arising from $0 \to A \otimes B \to A \Uparrow \otimes B \to A^* \otimes B \to 0$ (1.9.4), on the right. Commutativity is precisely the identity (ii) of 1.9.5.
