Big list: collecting results of diagram chasing

Question

I plan giving a homework problem on linear algebra for my students and my idea was to collect many results of diagram chasing.

Could you help me and give me some results of diagram chasing you know? They can be very basic or they can be more difficult. I think of results such as the "Five lemma".

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Remarks

PS: So far, they know all the basic stuff of linear algebra (vector space, linear maps, kernels, images, etc.). They don't know what is a finite-dimension space but they know what is a free/spanning finite families and finite basis.

PPS: They don’t know what is the quotient $E/F$... but maybe this can be overpassed by using subspaces $G$ such that $E = F \oplus G$.

Answer 1

The 3x3 lemma: it is usually proved via the Snake lemma but can be proved by direct diagram chasing.

Answer 2

Of course, there is the Snake lemma but it uses the cokernel.

Answer 3

One very classical result is the Five lemma.

Answer 4

There is also the Short five lemma.

Answer 5

It's a bit late, but here's another interesting one:

If in the short five lemma, you replace "is an isomorphism" by "is $0$", you get a wrong statement, but if you put two short exact sequences on top of one another and only ask for the composite to be $0$, then it becomes true. More precisely :

Suppose we have a commutative diagram $$\require{AMScd}\begin{CD}0 @>>>A @>>> B @>>> C @>>> 0\\ @VVV @VVV@VVV@VVV@VVV\\ 0 @>>>D @>>> E @>>> F @>>> 0\\ @VVV @VVV@VVV@VVV@VVV\\ 0 @>>>H @>>> K @>>> L @>>> 0\end{CD}$$ with exact lines, and suppose $A\to D, C\to F, D\to H$ and $F\to L$ are $0$. Then the composite $B\to E\to K$ is zero. Find an example where $B\to E$ isn't.

(Note : naively you would expect there to be no counterexamples if the short exact sequences split, so you would expect it to hold with just one map on vector spaces, but even then it doesn't)

It's an instructive diagram chasing argument, and finding a counterexample is instructive as well - moreover, it is related to deep mathematics.