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Suppose that you have the following diagram of modules over some ring: enter image description here

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

Thanks in advance.

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  • $\begingroup$ O.K. I will do that. $\endgroup$ – user28083 Dec 29 '15 at 17:28
  • $\begingroup$ 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. $\endgroup$ – Gregory Grant Dec 29 '15 at 17:37
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    $\begingroup$ "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. $\endgroup$ – JustAskin Dec 29 '15 at 17:48
  • $\begingroup$ @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)? $\endgroup$ – user28083 Dec 29 '15 at 18:09
  • $\begingroup$ 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. $\endgroup$ – Mark Bennet Dec 29 '15 at 18:33
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$sr=tu$ 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 $gf=kh.$

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  • $\begingroup$ I think to prove your statement you need an actual counter-example. $\endgroup$ – Gregory Grant Dec 29 '15 at 17:38
  • $\begingroup$ @Justpassingby: what about part one of the question? $\endgroup$ – user28083 Dec 29 '15 at 17:48
  • $\begingroup$ I think my first paragraph answers that. Your five conditions are insufficient, you need to include $sr=tu.$ $\endgroup$ – Justpassingby Dec 29 '15 at 17:50
  • $\begingroup$ @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)? $\endgroup$ – user28083 Dec 29 '15 at 18:16
  • $\begingroup$ Unless there are other hypotheses, those authors are incorrect. The counterexamples offered in my first paragraph and in other comments and answers demonstrate that. $\endgroup$ – Justpassingby Dec 29 '15 at 19:29
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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.

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