Weibel 1.5.8: Well-definedness of long exact sequence containing homology of $f$

Question

I'm currently reading Weibel's An Introduction to Homological Algebra and I am a little stuck in section 1.5.8. In the sections before it is proven that the homology of each map $f : B_. \to C_.$ occurs in some long exact sequence, using mapping cones and cylinders. Namely, the homology of $f$ occurs in the long exact sequences obtained from the complexes $$0 \to B_. \to \text{cyl}(f) \to \text{cone}(f) \to 0 \qquad \text{and} \qquad 0 \to C_. \to \text{cone}(f) \to B[-1] \to 0.$$

He next claims that "the long exact sequence is well defined". I'm not completely sure what is meant by this formulation. I suppose, it could mean that up to some nice equivalence relation (isomorphism maybe?), there is only one such long exact sequence. The only thing proven in this section, however, is that for a monomorphism $f$ the long exact sequence obtained from $\text{cone}(f)$ and $\text{cyl}(f)$ is isomorphic to the one obtained from the short exact sequence $$0 \to B_. \stackrel{f}{\to} C_. \stackrel{g}{\to} D_. \to 0.$$ Here $g$ is the cokernel of $f$. An exercise provides the dual statement for epimorphisms. I don't really see a connection to the statement we want to prove here.

My questions are:

1. Is my interpretation of the phrase "the long exact sequence is well defined" correct?
2. How do the results about monos and epis prove this?

Answer 1

To my understanding, the main point of that section is the following:

So up to this point, we know how to fit the homology of $f : B_{\bullet} \to C_\bullet $ into some long exact sequence by either using one of the short exact sequences:

$$ 0 \to B_\bullet \to \text{cyl}(f) \to \text{cone}(f) \to 0 \ (*) \qquad \text{and} \qquad 0 \to C_\bullet \to \text{cone}(f) \to B[-1] \to 0 \ (**).$$

Now we want to show that in fact, this process of fitting the homology is "unique" (up to quasi-isomorphism), meaning, if we have any other $D_\bullet$ such that $0 \to B_\bullet \stackrel{f}{\to} C_\bullet \stackrel{g}{\to} D_\bullet \to 0$ is exact, then the long exact sequence that it produces must be the "same" (up to quasi-isomorphism) as the long exact sequence that we get from applying $(*)$ or $(**)$.

The proof was shown in the book or explained more explicitly by butter-imbiber above (but I think you already knew). I'm not sure I understand the part when you mentioned about monos and epis $f$ though, since it was proved for any $f$ in the textbook the result that I mentioned above.

Answer 2

The answer to Q1 is yes. I'll expand on an answer to Q2 below.

(I don't know how to get tikzcd diagrams onto StackExchange and I don't have enough reputation to include images. You'll have to follow these links to see what I mean. I also don't have enough reputation to comment. My apologies.)

In 1.5.8, Weibel draws a diagram with exact rows in the category of complexes. Let's call this diagram (1).

The three rows in (1) have been seen before in 1.5.2 and on p22. The vertical morphisms are $\alpha$, $\beta$ and $\varphi$. The morphism $\varphi$ is new, but $\alpha$ was seen in Lemma 1.5.6 (where, confusingly, the Lemma uses $C$ while (1) uses $B$) and $\beta$ was seen just afterwards in Exercise 1.5.4. We know both $\alpha$ and $\beta$ are quasiisomorphisms. We can deduce that $\varphi$ is a quasiisomorphism too, by the argument Weibel gives after (1) on p23.

Applying homology to each of the three rows in this diagram gives us the rows in the following diagram, all of which are long exact sequences. We'll call this second diagram (2). The vertical maps in diagram (2) are either identities (which are isomorphisms) or the morphisms $H_n(\alpha),H_n(\beta),H_n(\varphi)$ (which are isomorphisms).

All the vertical morphisms in (2) are therefore all isomorphisms. This proves that all three long exact sequences are isomorphic.

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EDITS: copy editing