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I got a book on homological algebra in a textbook giveaway and I'm just starting to learn more about exact sequences in preparation for reading the book more seriously. I have seen things like the following before but don't understand them:

These lemmas, in my limited understanding, show you how to propagate exactness conditions through a commutative diagram or infer additional properties of some morphism somewhere.

These lemmas are hard to appreciate, though, without some examples in mind of what can go wrong.

So, I'm wondering if there are any instructive examples of fake diagram lemmas that are wrong for a subtle reason.

Ari Brodsky
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Greg Nisbet
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    To me, the amazing thing about the snake lemma is not the exactness, but rather that there is any (non-trivial) natural map at all $\mathrm{Ker}(c)\to \mathrm{Coker}(a)$. – Alex Kruckman Dec 26 '23 at 19:18
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    For the five lemma and nine lemma, why not try to weaken the hypotheses (I.e. assume less about the maps in the diagram) and look for counterexamples to these weakened statements? – Alex Kruckman Dec 26 '23 at 19:21
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    "Instructive" and "subtle" are rather subjective. – John Palmieri Dec 26 '23 at 21:54
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    @JohnPalmieri Yes and for this reason I have voted to close. The question is opinion-based, and not really focussed either. But your example below is really nice, btw. – Martin Brandenburg Dec 26 '23 at 21:56
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    @MartinBrandenburg I think the question has some merit, if viewed as "what are some fake diagram lemmas?" I agree that some more focus would help. – John Palmieri Dec 26 '23 at 22:13
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    @MartinBrandenburg How about this: most disciplines in mathematics have their "cautionary tales", e.g. things like the topologist's sine curve. Such tales are instructive examples of the requirement for as much rigour as we can manage... so it would be instructive to have cautionary tales in diagram chasing! – FShrike Dec 26 '23 at 22:22
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    See this question for an instructive fake diagram lemma: the false case of the nine lemma. – Misha Lavrov Dec 27 '23 at 14:32

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If "fake" diagram lemmas fail, it likely won't be for a subtle reason, but rather just because there is a counterexample.

Anyway, here is one example: given a commutative diagram of exact sequences $$ \require{AMScd} \begin{CD} 0 @>>> A @>>> B @>>> C @>>> 0 \\ @. @V{0}VV @V{f}VV @V{0}VV\\ 0 @>>> A' @>>> B' @>>> C' @>>> 0 \\ \end{CD} $$ it need not be the case that $f=0$.

John Palmieri
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    For anyone wondering, the diagram tells you $f$ vanishes on $A\subseteq B$ and the fact $C\to C'$ is zero only tells you that $f(B)\subseteq A'$; being zero "on homology" doesn't entail being literally equal to zero. A simple example is provided by the short exact sequence of Abelian groups $0\to\Bbb Z/2\Bbb Z\hookrightarrow\Bbb Z/4\Bbb Z\twoheadrightarrow\Bbb Z/2\Bbb Z\to0$ and the endomorphism $(0,f,0)$ of this sequence, where $f$ is multiplication by $2$. – FShrike Dec 26 '23 at 22:18
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    If the whole picture is an endomorphism ($A=A'$, $B=B'$, $C=C'$), then you can prove that $f \circ f = 0$. – John Palmieri Dec 26 '23 at 22:41
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    @FShrike Another example is to just take $A$ and $C'$ to be $0$, and let $A'=B'=B=C$, with identity maps at the appropriate places in the diagram above. Incidentally, this shows that the endomorphism requirement in John's comment is necessary. – Daniël Apol Dec 26 '23 at 22:43