Consider Lemma 1 from Beilinson's paper "Coherent Sheaves on $\mathbb{P}^n$ and Problems of Linear Algebra", as follows.

Let $\mathcal{C}$ and $\mathcal{D}$ be triangulated categories, $F: \mathcal{C} \to \mathcal{D}$ an exact functor, $\{X_i\}$ be a family of objects of $\mathcal{C}$. Let us assume that $\{X_i\}$ generates $\mathcal{C}$, $\{F(X_i)\}$ generates $\mathcal{D}$, and for any pair $X_i$, $X_j$ from the family $F: \text{Hom}^\bullet (X_i , X_j) \to \text{Hom}^\bullet(F(X_i), F(X_j))$ is an isomorphism. Then $F$ is an equivalence of categories.

My question is, what is the correct way to think about it/the intuition for this? Thanks.

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    $\begingroup$ Have you tried proving the lemma? I think the data of the $X_i, \text{Hom}^\bullet(X_i, X_j)$ should be though of as "generators and relations" for $\mathcal{C}$, and the assumption is that $\mathcal{D}$ has an isomorphic set of "generators and relations." But the proof is easy enough to be considered intuitive, in my opinion... $\endgroup$ – Daniel Litt Oct 13 '15 at 23:01
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    $\begingroup$ I'm really not sure what you want. The statement involves various somewhat technical notions, and without knowing your "intuitive" understanding of these, it seems impossible to give an "intuitive" explanation of why the lemma is true. I assume from the way you asked the question that you know a proof. Maybe you could say what you find "unintuitive" about that proof. $\endgroup$ – Jeremy Rickard Oct 17 '15 at 12:55
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    $\begingroup$ I agree with @JeremyRickard's comment. What, in your understanding or point of view, counts as an "intuition" behind a given result? $\endgroup$ – Yemon Choi Nov 2 '15 at 1:55
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    $\begingroup$ There have been several similar questions recently, e.g. mathoverflow.net/questions/215847/… and mathoverflow.net/questions/215665/… They appear to be very popular. I must say that I think "[Theorem statement] What is the intuition behind this theorem?" is not a good question template. $\endgroup$ – Lazzaro Campeotti Nov 2 '15 at 10:45
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    $\begingroup$ ... and therefore the whole of $\mathcal{D}$. It's a routine application of the definition of "generates" and I'm not really sure how I'd describe the "intuition" for it, other than saying "a triangulated subcategory containing a generating set is the whole category". But that's just restating the definition of "generates", not really giving any particular intuition, as far as I see. $\endgroup$ – Jeremy Rickard Nov 2 '15 at 14:29

I think some of the commenters have forgotten the time when they found vector space linear algebra understandable, but triangulated categories confusing.

For someone in such a state, a useful tool to help understand statements about triangulated categories is passage to the Grothendieck group. Recall that this is done by taking the abelian group generated by the objects in the category $\mathcal{C}$, divided by the relation in which the sum of three elements of a triangle is zero. This gives an abelian group, and I'll tensor it with a field to get a vector space $[\mathcal{C}]$. Supposing in addition that Hom in the category gives finite dimensional vector spaces, $[\mathcal{C}]$ gets a bilinear form by $B(X, Y) = \dim Hom(X, Y)$.

For instance, in this case the thus "decategorified" statement is as follows:

Let $C$ and $D$ be vector spaces equipped with bilinear forms, and $F: C \to D$ a linear map. Say $X_i$ is a spanning set of vectors for $C$ and $F(X_i)$ spans $D$. Suppose $F$ is "an isometry on the spanning set", i.e., $B(X_i, X_j) = B(F(X_i), F(X_j))$. Then $F$ is an isometry.

This is a statement more accessible to intuition. And, as a proof of the original statement necessarily "decategorifies" to a proof of the decategorified one, often one can proceed in reverse and first prove the decategorified statement and then try to lift.

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    $\begingroup$ Some slight comments--perhaps to define the Grothendieck group you want to set the alternating sum of objects in a distinguished triangle to zero? And it might be worth remarking that the $\dim$ in your definition of $B$ is taken in the graded sense, and $\text{Hom}$ should be $\text{Hom}^\bullet$. But I like this answer a lot! $\endgroup$ – Daniel Litt Nov 5 '15 at 0:42
  • $\begingroup$ This was exactly what I was looking for, thanks so much! $\endgroup$ – user61522 Nov 5 '15 at 3:35

For the sake of completeness, I will include some comments on the "decategorified" statement; hopefully they are not too elementary as to be considered useless.

Let $C$ and $D$ be vector spaces equipped with bilinear forms, and $F: C \to D$ a linear map. Say $X_i$ is a spanning set of vectors for $C$ and $F(X_i)$ spans $D$. Suppose $F$ is "an isometry on the spanning set", i.e., $B(X_i, X_j) = B(F(X_i), F(X_j))$. Then $F$ is an isometry.

Our forms are bilinear, so intuitively, this is just saying that if we have a linear map $F:C \to D$, then preserving the bilinear form on the basis vectors is enough to guarantee that it preserves the bilinear form on the entire space.

This is just because of the linearity. A linear map is completely determined by where it sends the basis vectors of the preimage. And the forms we want to preserve are also bilinear. So if we look at $B_{1}(u,v)$, with $u$ and $v$ in $C$, we can write$$u = a_{1}X_{1} + \ldots + a_{n}X_{n},\text{ }v = b_{1}X_{1} + \ldots + b_{n}X_{n}.$$ Then $$B_{2}(F(u),F(v)) = \sum_{i}\sum_{j} a_{i}b_{j}B_{2}(F(X_{i}),F(X_{j})) = \sum_{i}\sum_{j} a_{i}b_{j}B_{1}(X_{i},X_{j}) = B_{1}(u,v).$$ Note that we have two different bilinear forms, $B_{1}$ on $C$ and $B_{2}$ on $D$.


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