Checking commutativity of a diagram of modules over some ring and what the commutativity of the diagram implies.

Suppose that you have the following diagram of modules over some ring:

These are my questions:

(1) To prove that the diagram is commutative, we needs to prove that $gf=kh$, $wf=rv$, $zh=uv$, $sw=xg$, $xk=tz$, am I right?

(2) Suppose that the above diagram is commutative. Do the following equalities hold: $sr=tu$, $swf=xkh$

The multiplication above means composition of maps

• O.K. I will do that. – user28083 Dec 29 '15 at 17:28
• What if $E,G,H$ all equal $\Bbb Z$ with $u,t$ the identity map, and all other rings are the zero ring. I think that might be a counter-example. – Gregory Grant Dec 29 '15 at 17:37
• "Suppose that the above diagram is commutative. Do the following equalities hold: $sr=tu$, $swf=xkh$" - The definition of a diagram being commutative is that all directed paths with the same endpoints are equal, so, yes. – JustAskin Dec 29 '15 at 17:48
• @Justaskin: Do commutativity of the above diagram imply also that $sr=tu$? And if that is the case, why do authors check commutativity by considering only the squares as I did in part(1) (See Gregory Grant answer)? – user28083 Dec 29 '15 at 18:09
• You have $tuv=tzh=xkh=xgf=swf=srv$ but you have nothing which says you can cancel $v$ if you only have the relations in $(1)$. Note that $sr=tu$ is a relation based on a square - you have missed the square $EFGH$, as you would see if you looked at the diagram as the projected faces of a cube - one relation for each face makes six. – Mark Bennet Dec 29 '15 at 18:33

$sr=tu$ needs to be included in the conditions, it does not follow automatically. To see this let all the other maps be 0.
$swf=xkh$ follows from $sw=xg$ and $gf=kh.$
• I think my first paragraph answers that. Your five conditions are insufficient, you need to include $sr=tu.$ – Justpassingby Dec 29 '15 at 17:50
• @Justpassingby: Do commutativity of the above diagram imply also that $sr=tu$? And if that is the case, why do authors check commutativity by considering only the squares as I did in part(1)? – user28083 Dec 29 '15 at 18:16
What if $E,G,H$ all equal $\Bbb Z$ with $u,t$ the identity map, and all other rings are the zero ring. I think that might be a counter-example.