At last we come to it: the greatest diagram lemma of all time.

The Salamander Lemma

For each piece of the double complex of the form

we obtain the following salamander-shaped diagram of mural maps:

Don't think this looks like a salamander? Fair enough. How about if we adjust it a bit ...

Still doesn't look like a salamander? One more try:

Okay, it's definitely a stretch. But I didn't name this lemma, so don't blame me! In any case, here's the (in)famous lemma:

I'll include the full proof of this at some point, but for now here is an official reference. Notice how it's not very long!

To assuage my guilt, I'll include at least a part of the proof. This is really one-eighth of the full proof, although all eight parts are very similar.

Suppose the diagram below is part of a double complex:

By composing the extramural map $A_{◻}{\to }^{◻}B$ (which is given on representatives by ${\partial }^{\text{ver}}$) with the intramural map ${}^{◻}B\to B^{\text{hor}}$ (which is the identity on coset representatives), we obtain a map $A_{◻}\to B^{\text{hor}}$. Composing that map with the intramural map $B^{\text{hor}}\to B_{◻}$, we obtain the 3-term sequence

For absolute clarity, we note that if $[a]\in A_{◻}$, then its image under the first map is $[{\partial }^{\text{vert}}(a)]\in B^{\text{hor}}$, and the image under the second map is $[{\partial }^{\text{vert}}(a)]\in B_{◻}$. The bracket notation for cosets is hiding the different types of cosets involved in these maps.

In any case, let's prove this sequence is exact at $B^{\text{hor}}$. First suppose an element $[b]\in B^{\text{hor}}$ is in the kernel of the intramural map $B^{\text{hor}}\to B_{◻}$. Recall that this intramural map is simply the identity on coset representatives, so we are now assuming the coset $[b]\in B_{◻}$ is the zero coset. By the definition of the donor, this implies $b$ is the sum of elements in the images of ${\partial }^{\text{vert}}$ and ${\partial }^{\text{hor}}$ mapping to $B$. In other words, there are elements $a\in A$ and $e\in E$ (where $E\to B$ is the map to the left of $B$) such that $b={\partial }^{\text{vert}}(a)+{\partial }^{\text{hor}}(e)$.

Next, recall that we began with an element $[b]\in B^{\text{hor}}$. These coset representatives are only defined up to elements in the image of $E\to B$, so $[b]=[{\partial }^{\text{vert}}(a)]$.

Now notice that the element $b\in \ker ({\partial }^{\text{hor}})$, so applying ${\partial }^{\text{hor}}$ to the previous equation gives

The last term on the right vanishes (since our diagram is part of a double-complex), so we can conclude ${\partial }^{\text{hor}}{\partial }^{\text{vert}}(a)=0$. In other words, $a\in \ker ({\partial }_{\text{out}}^{\text{diag}})$, and so $[a]\in A_{◻}$.

We've therefore proven $[b]\in B^{\text{hor}}$ is in the image of $A_{◻}\to B^{\text{hor}}$. So, we have shown that the kernel of $B^{\text{hor}}\to B_{◻}$ is contained in the image of $A_{◻}\to B^{\text{hor}}$.

Conversely, take any $[a]\in A_{◻}$. Then the image of $[a]$ under the map $A_{◻}\to B^{\text{hor}}$ is $[{\partial }^{\text{vert}}(a)]$. The element ${\partial }^{\text{vert}}(a)$ is (by definition!) in the image of ${\partial }^{\text{vert}}$ and hence is zero in the donor $B_{◻}$. Thus, the image of $A_{◻}\to B^{\text{hor}}$ is contained in the kernel of $B^{\text{hor}}\to B_{◻}$.

Extramural isomorphisms

Now let's see some corollaries of the Salamander Lemma. First let's focus on situations in which the mural maps are isomorphisms.

The above is an immediate corollary of the corollary of the Salamander Lemma. For instance, the first statement follows from the exactness of the sequence below:

Intramural isomorphisms

As for the intramural maps, we have the following results. If it looks intimidating, note that it's covering four very similar situations, which we've described by which "edge" of the double-complex we're considering.

Let's prove the above statement. First notice that we can extend the lower row to the left by inserting a zero object and arrow. Then, by our assumption, that bottom row is horizontally exact at $0$ and $B$:

By our extramural isomorphism them, we then have a horizontal isomorphism $0_{◻}{\simeq }^{◻}B$. This proves ${}^{◻}B\simeq 0$, which we will use shortly.

Next consider the following part of the double-complex, where the $\bullet$ indicates whatever object happens to be in that position. (Remember that we can always add zero objects and zero arrows, if necessary.)

By the Salamander Lemma, we have a six-term exact sequence

Since $0_{◻}{\simeq }^{◻}B\simeq 0$, the end of this exact sequence is the 4-term exact sequence

This proves ${}^{◻}A\simeq A^{\text{hor}}$.

Similarly, now consider the following part of the double-complex:

By the Salamander Lemma, we have a six-term exact sequence

Once again, since $0_{◻}{\simeq }^{◻}B\simeq 0$, the beginning of this exact sequence is the 4-term exact sequence

This proves $A^{\text{vert}}\simeq A_{◻}$.

The following three analogous situations have similar proofs:

Let's see how a multitude of famous "diagram lemmas" now all immediately follow.

Suggested next note

Diagram lemmas