Slightly simplified, the five lemma states that if we have a commutative diagram (in, say, an abelian category)

$$\require{AMScd} \begin{CD} A_1 @>>> A_2 @>>> A_3 @>>> A_4 @>>> A_5\\ @VVV @VVV @VVV @VVV @VVV\\ B_1 @>>> B_2 @>>> B_3 @>>> B_4 @>>> B_5 \end{CD} $$ where the rows are exact and the maps $A_i \to B_i$ are isomorphisms for $i=1,2,4,5$, then the middle map $A_3\to B_3$ is an isomorphism as well.

This lemma has been presented to me several times in slightly different contexts, yet the proof has always been the same technical diagram chase and no further intuition behind the statement was provided. So my question is: do you have some intuition when thinking about the five lemma? For instance, particular choices of the $A_i, B_i$ which make it more transparent why the result should be true? Some analogy, heuristic, ...?


One can think of the five lemma in terms of the two four lemmas. I think this makes it clearer... for instance drop the $A_1$ and $B_1$ from your diagram. If the maps from $A_2$ and $A_4$ to $B_2$ and $B_4$ are epimorphisms and the morphism $A_5 \to B_5$ is monic then the cokernel of $A_3 \to A_4$ is a subobject of the cokernel of $B_3 \to B_4$. So morally $B_3$ is an "extension" of quotients of $A_2$ and $A_4$ and we have not "killed less stuff" in the bottom row so $A_3 \to B_3$ should also be an epimorphism.


Since long exact sequences come from splicing together short exact sequences, you might as well worry about the case where $A_1=A_5=B_1=B_5=0$ (at least as far as intuition is concerned). This follows from the Snake Lemma, of course, but the version where the outside vertical maps are isomorphisms is an even easier diagram chase. Perhaps that will illuminate.

In terms of why it's true without chasing elements, think about the same simplified version, but just for Abelian groups. In general, of course, you can have $G/H \cong G'/H'$ for lots of groups $G,G'$ and respective subgroups $H,H'$. In general, that isomorphism won't even lift to a homomorphism $G \to G'$, much less an isomorphism. Similarly, you could have $H \cong H'$ without that isomorphism extending to a homomorphism $G \to G'$. If, however, you have both, then the attempt to lift the one isomorphism and the attempt to extend the other both succeed.

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    $\begingroup$ Thanks a lot for working on my intuition! Maybe I'm misreading your final statement but I think you need to assume the map between G and G' to exist from the beginning. Take eg. G=Z/4, G'=Z/2xZ/2 and factor out by Z/2 subgroups. $\endgroup$ – Armin Straub Oct 26 '09 at 22:57
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    $\begingroup$ Yes, you're right. Silly me. Still, for intuition's sake maybe it's close to the truth (he said, trying to save face). $\endgroup$ – Graham Leuschke Oct 26 '09 at 23:59

Coming back to Graham Leuschke's answer: Assume $A_1=A_5=B_1=B_5=0$. Your simplified version of the five lemma then follows because you can simply replace $A_2$ by $\mathrm{coker}(A_1\to A_2)$ etc. We may even use the isomorphisms given to identify $A_2=B_2$ and $A_4=B_4$.

Now, both $A_3$ and $B_3$ are extensions of $A_4=B_4$ by $A_2=B_2$. The simplified five lemma becomes: two extensions are isomorphic if and only if there exists a map between them that is compatible with the inclusion and the projection morphisms. You could now cook up a different proof by comparing group cocycles defining $A_3$ and $B_3$. Maybe this gives some intuition.


One example would be a map induced by a morphism $f: X \to Y$ in the long homology sequence.

E.g. suppose the top row is a cohomology of pair $(X, A)$ and the bottom row is the cohomology of pair $(Y, B)$. Then the theorem says that the $H^n(X, A)$ can be squeezed between the $n$-th and $(n-1)$-th cohomology of $X$ and $A$, because any morphism inducing isomorphism on those extends to $H^n(X, A)$.

  • $\begingroup$ Q: Where does the long homology sequence come from? A: From repeated application of the 5 Lemma or Snake Lemma. So perhaps not very illuminating. $\endgroup$ – Graham Leuschke Oct 26 '09 at 23:56
  • $\begingroup$ Duh! That's the statement about snake lemma being the technical equivalent of our intuition about cohmology. $\endgroup$ – Ilya Nikokoshev Oct 26 '09 at 23:59
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    $\begingroup$ The long exact sequences found in homotopy theory (hence in homological algebra) can be proved perfectly well from abstract homotopy theory without any knowledge of the five or snake lemmas. $\endgroup$ – Denis-Charles Cisinski Oct 27 '09 at 0:39

Variant of what other people already said, but I find the following statement quite intuitive:

  • If G is a group and $f: X \to Y$ is a $G$-equivariant map inducing a bijection on orbit sets $X/G \to Y/G$ and isomorphism of all stabilizer groups $G_x \to G_{f(x)}$, then $f$ must be a bijection.

Now in the setting of the 5-lemma, specialize to $X = A_3$, $Y = B_3$, and $G = A_2 \cong B_2$ acting by addition on $X$ and $Y$. By exactness of the rows, all maps of stabilizers are identified with the isomorphism $\mathrm{Im}(A_1 \to A_2) \cong \mathrm{Im}(B_1\to B_2)$ induced by the leftmost square in the diagram, and the map of orbit sets is identified with the isomorphism $\mathrm{Ker}(A_4 \to A_5) \cong \mathrm{Ker}(B_4 \to B_5)$ induced by the rightmost square.

From this point of view it is also clear that "abelian group" is more structure than necessary. For example, the rightmost square need only be given in the category of pointed sets.


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