Motivation for definitions of donor and receptor in Salamander Lemma?

Question

$\newcommand{\im}{\operatorname{Im}}$Consider the following (subpart of) a double complex, using the same notation as in George Bergman's pre-print or in these lecture notes:

$$\require{AMScd}\begin{CD} \bullet @>a>> \bullet @>>> \bullet \\ @VbVV @VcVV @VVV \\ \bullet @>d>> A @>e>> \bullet \\ @VVV @VfVV @VgVV \\ \bullet @>>> \bullet @>h>> \bullet \end{CD} $$

and define $p = c\circ a = d\circ b$ (diagonal into $A$), $q= g \circ e = h \circ f$ (diagonal out of $A$).

Question: What "information" is contained in the modules $\dfrac{\ker (q)}{\ker (e) + \ker(f)}$ and $\dfrac{\im(c) \cap \im(d)}{\im(p)}$ which is interesting/is useful/we care about?

Here is what I believe to be an equivalent question:

Question: (i) Why is the donor of $A$ defined as $\dfrac{\ker (q)}{\im(c) + \im(d)}$ instead of $\dfrac{\ker(e) + \ker(f)}{\im(c) + \im(d)}$?
(ii) Why is the receptor of $A$ defined as $\dfrac{\ker(e) \cap\ker(f)}{\im (p)}$ instead of $\dfrac{\ker(e) \cap \ker(f)}{\im(c) \cap \im(d)}$?

(Probably unnecessary) Background: Since one has that $\im (\psi \circ \phi) \subseteq \im(\psi)$ and $\ker(phi) \subseteq \ker(\psi \circ \phi)$ for any morphisms $\psi, \phi$, one always has the chains of inclusion: $$ \im(p) \subseteq \im(c) \cap \im(d) \subseteq \ker(e) \cap \ker(f) $$ and $$ \im(c) + \im(d) \subseteq \ker(e) + \ker(f) \subseteq \ker(q)$$ (since of course that it's a double complex means that $\im(c) \subseteq \ker(f)$ and $\im(d) \subseteq \ker(e)$).

Denote by $D = D_1/D_2$ (with $D_1 \subseteq A \supseteq D_2$) a "candidate donor" and $R = R_1 /R_2$ (with $R_1 \subseteq A \supseteq R_2$) a "candidate receptor". Denote by $A^v$ the vertical homology $\ker (f) / \im(c)$ and denote by $A^h$ the horizontal homology $\ker(e) / \im(d)$. Then consider the following properties we want to be satisfied by any candidate donor or receptor.

Intramural properties: (Compare Lemma 1.2 or Lemma 2.)

• Inclusion $R_1 \to \ker(f)$ and $R_1 \to \ker(e)$ should induce morphisms $R \to A^v$ and $R \to A^h$.
• Inclusion $\ker(f)\! \to D_1$ and $\ker(e) \to D_1$ should induce morphisms $A^v \to D$ and $A^h \to D.$

Extramural Properties: (Compare Lemma 1.4 or Lemma 3.)

Letting $D_{(A)}$ denote the (candidate) donor of $A$, and analogously $R_{(B)}$ the (candidate) receptor of $B$, then a morphism (horizontal or vertical) in the double complex $A \to B$ should induce a morphism $D_{(A)} \to R_{(B)}$. (According to George Bergman's pre-print this property is the source of/motivation for the names "donor" and "receptor".)

Assuming I am not making any overly stupid errors, the intramural properties should require (compare the comment in Anton Geraschenko's blog post) that:

• $\{0\} \subseteq R_1 \subseteq \ker(e) \cap \ker(f)$.
• $\{0\} \subseteq R_2 \subseteq \im(c) \cap \im(d)$.
• $\ker(e) + \ker(f) \subseteq D_1 \subseteq A$.
• $\im(c) + \im(d) \subseteq D_2 \subseteq A$.

Moreover, the obvious requirements that $R_2 \subseteq R_1$ and $D_2 \subseteq D_1$ suggest we want $R_1$ as large as possible and $D_2$ as small as possible, since the "upper bounds" for $R_1$ and $R_2$ will coincide when the complex is exact at $A$ horizontally and vertically (and similarly the "lower bounds" for $D_1$ and $D_2$ will also coincide when the complex is exact at $A$ horizontally and vertically).

So as far as I can tell, and obviously the above argument is very heuristic:

• any candidate donor $D$ should be of the form $\dfrac{D_1}{\im (c) + \im (d)}$.
• any candidate receptor $R$ should be of the form $\dfrac{\ker(e) \cap \ker(f)}{R_2}$.

Now, again barring any stupid algebra mistakes, it seems that any of the possible choices:

• $\ker(e) + \ker(f) \subseteq D_1 \subseteq A$,
• $\{0\} \subseteq R_2 \subseteq \im(c) \cap \im(d)$,

will (at least given the choices made for $D_2$ and $R_1$ above) satisfy the extramural properties due to the fact that it's a double complex and any two consecutive arrows equal the $0$ morphism.

So it's very unclear to me why one should prefer one of these choices of $D_1$ or $R_2$ to any other possible choice. As an arbitrary thing, it seemed like (since the choices of $A$ and $\{0\}$ seem to be very boring), the simplest thing would be to choose the smallest possible $D_1$ and the largest possible $R_2$, thus $D_1 = \ker(e) + \ker(f)$ and $R_2 = \im(c) \cap \im(d)$.

But apparently the correct thing to do is to choose $D_1 = \ker(q)$ and $R_2 = \im(p)$, and I can't figure out why. Why do we care about the "information" in $\ker(q) / (\ker(e) + \ker(f) )$ but not about the information in $A / \ker(q)$, such that we're not willing to discard the former but are willing to discard the latter? Similarly, why do we care about the information in $(\im(c) \cap \im(d))/\im (p)$, but not about the information in $\im(p)$, such that we're not willing to discard the former but willing to discard the latter? It seems like there must be another property of the donor/receptor which I am overlooking in order for this to be unclear to me.

Answer 1

Too long for a comment.

Heuristics. This section might be paraphrasing what you wrote, but since you asked for motivation, it might still give a helpful perspective. In my view, there are essentially only two candidate definitions for the notion of receptor and donor, respectively. Let me explain. Given a double complex in an abelian category, pick any object $A$ in it. Now, the double complex "induces subobjects of $A$". Namely, the kernels of all morphisms going out of $A$ and the images of all morphisms going into $A$ yield subobjects of $A$. Additionally, you can form the intersection/meet (the product in the posetal category of subobjects of $A$) and the union/join (the coproduct in said posetal category) of two such induced subobjects. Since the composition of two horizontal or of two vertical differentials is zero, the only part of the double complex that contains morphisms that induce interesting subobjects of $A$ is the following:

$$\require{AMScd}\begin{CD} \bullet @>a>> \bullet @>>> \bullet \\ @VbVV @VcVV @VVV \\ \bullet @>d>> A @>e>> \bullet \\ @VVV @VfVV @VgVV \\ \bullet @>>> \bullet @>h>> \bullet \end{CD} $$

There are three subobjects of $A$ induced by morphisms into $A$, namely $\operatorname{im}(c)$, $\operatorname{im}(d)$ and $\operatorname{im}(c \circ a)=\operatorname{im}(d \circ b)$. The sublattice of subobjects of $A$ generated by these three objects looks as follows:

$\hskip2in$

Analogously, one obtains the following lattice of subobjects by taking kernels and closing under join and meet:

$\hskip2in$

As you note, by definition of a double complex, we additionally have the inclusions $\operatorname{im}(c)\cap \operatorname{im}(d)\subseteq \operatorname{ker}(f)\cap \operatorname{im}(e)$, $\operatorname{im}(c) \subseteq \operatorname{ker}(f)$, $\operatorname{im}(d) \subseteq \operatorname{ker}(e)$ and $\operatorname{im}(c)+\operatorname{im}(d)\subseteq \operatorname{ker}(f)+\operatorname{im}(e)$. I visualize the two 5-element lattices stacked on top of each other with these four inclusion arrows between them. This yields a commutative diagram with ten vertices. From this commutative diagram it is immediate that by forming quotients (geometrically: compressing the two diagrams/lattices into one commutative diagram), the way Bergman does, we obtain the diamond-shaped diagram of intramural maps (Bergman's Lemma 1.2). More generally, pick two subobjects $\ker(e) + \ker(f) \subseteq D\subseteq A$ and $\{0\} \subseteq R\subseteq \operatorname{im}(c) \cap \operatorname{im}(d)$ and define $${^\Box}A:=\dfrac{\operatorname{ker}(f)\cap \operatorname{ker}(e)}{R} \hskip0.5in \text{and} \hskip0.5in A_{\Box}:=\dfrac{D}{\operatorname{im}(c)+\operatorname{ker}(d)}.$$ From the last commutative diagram with ten vertices (the two lattices stacked on top of each other) it follows that, as you note, even with these definitions Bergman's Lemma 1.2. holds. Now, there are only two natural candidates (we want to use subobjects of $A$ induced from the double complex) for $D$ and $R$, respectively. For $R$ we could either choose $\operatorname{im}(c)\cap\operatorname{im}(d)$ or $\operatorname{im}(c\circ a)=\operatorname{im}(d\circ b)$. For $D$ we could either choose $\operatorname{ker}(f)+\operatorname{ker}(e)$ or $\operatorname{ker}(h\circ f)=\operatorname{ker}(g\circ e)$. This gives four possible definitions for a donor-receptor pair. Any of these four definitions additionally satisfies the extramural properties (Lemma 1.3).

Why Bergman’s definition? Since Lemmas 1.2. and 1.3. hold for any of these definitions, we can formulate the salamander lemma with any of them. However, for all but Bergman's definition the salamander lemma fails. And isn't the whole point of the notion of receptors and donors that they make the salamander lemma true?

With any of the four possible definitions both salamander sequences are still chain complexes, but they are in general not exact. Consider, for instance, the following commutative diagram

$$\require{AMScd}\begin{CD} 0 @>>> 0 @>>> 2\mathbb{Z}\\ @VVV @VVV @ViVV \\ 0@>>> C=\mathbb{Z}@>1>>\mathbb{Z} \\ @VVV @V\begin{pmatrix}0\\1\end{pmatrix}VV @VpVV\\ \mathbb{Z} @>f>> A=\mathbb{Z}\oplus\mathbb{Z} @>g>> B=\mathbb{Z}/2\mathbb{Z}\\ @V1VV @V\begin{pmatrix}1&0\end{pmatrix}VV @VVV \\ \mathbb{Z} @>1>> \mathbb{Z} @>>>D=0 \end{CD}$$ where the map $i\colon 2\mathbb{Z}\rightarrow \mathbb{Z}$ is the canonical injection, the map $p\colon \mathbb{Z}\rightarrow \mathbb{Z}/2\mathbb{Z}$ is the canonical projection, and we have $f=\begin{pmatrix}1\\0\end{pmatrix}$ and $g=p\circ \begin{pmatrix}0&1\end{pmatrix}$. Extend this diagram to a double complex by letting every other entry be zero.

With your suggested definition, this complex yields the (non-exact) salamander sequence $$0\rightarrow 2\mathbb{Z}\rightarrow 0\rightarrow 0\ldots \rightarrow 0.$$