Draft lesson.

Split triangles, idempotent telescopes and semisimple structures

Original text public domain (CC0).

A distinguished triangle need not be a short exact sequence in its underlying category. Nevertheless, a categorical monomorphism in a triangulated category always splits. This makes an abelian triangulated category semisimple. We prove these assertions with the actual maps, construct the ordinary colimit of an idempotent tower from its telescope, and verify every triangulated axiom for a semisimple abelian category equipped with a translation equivalence.

Use the universe and Choice convention of Universes and small categories. A translation \(T\) is an additive equivalence with chosen coherent inverse \(S\), unit \(\eta:1\to ST\), and counit \(\epsilon:TS\to1\). Integer powers and the comparison maps in Translations, coherent powers and universal triangle comparisons, Sections 1–3 identify inverse translates explicitly. Our rotation convention is \((f,g,h)\mapsto(g,h,-Tf)\). We use the complete exact Hom proofs in Triangles, finite universal maps and the limits of cohomology, Lemma 1.1, and the finite sum and comparison proofs in Translations, Sections 6–7.

1. Four-term recognition with the missing composite supplied

Theorem 1.1. Suppose we have a commuting comparison of entire triangles

The upper triangle is distinguished, \(\alpha,\beta\) are isomorphisms, and

If either of the following sequences is exact at its two interior terms for every object \(P\), the lower triangle is distinguished:

Exactness of the second row alone suffices; exactness of both rows is unnecessary.

Proof for the first row. Fix \(P\). If \(z:P\to Z\) satisfies \(\gamma z=0\), the last square and invertibility of \(T\alpha\) give \(hz=0\). Upper Hom exactness gives \(z=gy\). Now \(g'\beta y=0\), so lower exactness at \(\operatorname{Hom}(P,Y')\) gives \(\beta y=f'x'\). Set \(x=\alpha^{-1}x'\). The first square gives \(\beta fx=f'x'=\beta y\), hence \(y=fx\) and \(z=0\).

For surjectivity, take \(z':P\to Z'\). The map \(t=(T\alpha)^{-1}h'z'\) is killed by \(Tf\): the translated first square and (1.2) give this actual zero composite. Upper exactness at \(\operatorname{Hom}(P,TX)\) supplies \(z:P\to Z\) with \(hz=t\). The last square then gives \(h'(z'-\gamma z)=0\). Lower exactness at \(\operatorname{Hom}(P,Z')\) gives \(z'-\gamma z=g'y'\). Thus

Every \(\operatorname{Hom}(P,\gamma)\) is bijective. Surjectivity at \(P=Z'\) supplies \(r:Z'\to Z\) with \(\gamma r=1\). Injectivity at \(P=Z\), applied to \(r\gamma\) and \(1\), gives \(r\gamma=1\).

Proof for the second row. We prove bijectivity of precomposition

If \(a'\gamma=0\), then \(a'g'\beta=0\), so \(a'g'=0\). Lower exactness supplies \(b':TX'\to P\) with \(a'=b'h'\). Then \(b'(T\alpha)h=0\). Upper exactness supplies \(c:TY\to P\) with \(b'(T\alpha)=c(Tf)\), absorbing the rotation sign into \(c\). The translated first square implies

For a map \(a:Z\to P\), the map \(ag\beta^{-1}:Y'\to P\) kills \(f'\). Lower exactness gives \(a'_0:Z'\to P\) with \(a'_0g'=ag\beta^{-1}\). Hence \((a-a'_0\gamma)g=0\); upper exactness gives \(b:TX\to P\) with \(a-a'_0\gamma=bh\). The required preimage is

Surjectivity in (1.5) at \(P=Z\) gives \(r\gamma=1\). Injectivity at \(P=Z'\) gives \(\gamma r=1\). Thus \(\gamma\) is an isomorphism in either case. All three vertex maps in (1.1) are isomorphisms and preserve the three arrows. Isomorphism closure of distinguished triangles proves the assertion. \(\square\)

The extra condition (1.2) has a precise role: it supplies the zero composite needed at the fifth Hom position. Merely calling a four-term comparison a five-term comparison would omit that step.

2. Zero cones and actual split triangles

Proposition 2.1. In a distinguished triangle

the map \(f\) is invertible if and only if \(Z\) is a zero object.

Proof. Compare the identity triangle \(X\xrightarrow1X\to0\to TX\) with (2.1), using \(1_X,f\) on its first two vertices. Square completion supplies the third map. If \(f\) is invertible, the two-isomorphism comparison theorem makes \(0\to Z\) invertible. Conversely, if \(Z\) is zero, that theorem applied to the first and third vertices makes \(f\) invertible. The theorem's complete proof uses exact five-term Hom rows and constructs the Yoneda inverse, so neither direction assumes a categorical cone functor. \(\square\)

Theorem 2.2. For (2.1), the following are equivalent:

1. \(h=0\).
2. There is \(s:Z\to Y\) with \(gs=1_Z\).
3. There is \(k:Y\to X\) with \(kf=1_X\).
4. The entire triangle is isomorphic, fixing its endpoints, to

For every section \(s\) in assertion 2, the specified middle isomorphism is \([f,s]:X\oplus Z\to Y\).

Proof. If \(h=0\), exactness of \(\operatorname{Hom}(Z,-)\) lifts \(1_Z\) through \(g\), producing \(s\). Conversely \(gs=1\) implies \(h=hgs=0\). If \(kf=1\), the zero composite \((Tf)h=0\) gives \(h=(Tk)(Tf)h=0\).

When \(h=0\), its inverse translate is also zero. For every \(P\), exact Hom therefore gives the split short exact sequence

The splitting is \(s_*\). Consequently \(\operatorname{Hom}(P,[f,s])\) is bijective: for \(y:P\to Y\), subtract \(sgy\), lift the remainder uniquely through \(f\), and use \(g\) to recover the second component. The Yoneda inverse construction in Theorem 1.1 makes \([f,s]\) invertible. Its first square is the definition, its second is \(gf=0,gs=1\), and the last arrows are zero. This proves assertion 4. Put \(k=\pi_X[f,s]^{-1}\) to obtain assertion 3. Finally (2.2) is distinguished: it is the finite direct sum of the identity triangle of \(X\) and the rotated identity triangle \(0\to Z\xrightarrow1Z\to0\). The finite-sum theorem includes the actual translation comparison on the last map. \(\square\)

Corollary 2.3. Every categorical monomorphism in a triangulated category is a split injection into a biproduct. Every categorical epimorphism is a split projection.

Proof. Complete a monomorphism \(f:X\to Y\) to (2.1). In the inverse rotation, the preceding map \(a:SZ\to X\) satisfies \(fa=0\). Left cancellation gives \(a=0\); the inverse-translation comparison identifies its translate, up to sign and invertible comparison, with \(h\), so \(h=0\). Theorem 2.2 gives \(Y\cong X\oplus Z\) and identifies \(f\) with the canonical injection.

For an epimorphism \(g:Y\to Z\) in (2.1), \(hg=0\) and right cancellation give \(h=0\), so \(g\) is the canonical projection in the same decomposition. Any specified epimorphism can be made the second arrow by completing it and inversely rotating. \(\square\)

These are categorical cancellation statements. A map that is injective only after applying one cohomology functor need not split in the original category.

3. The signed triangulated translation

Proposition 3.1. Translation is a triangulated equivalence with natural comparison

Thus its image of (2.1) has actual last arrow \(-Th\).

Proof. Three rotations of (2.1) give the distinguished triangle with arrows \((-Tf,-Tg,-Th)\). Vertex maps \((1_{TX},-1_{TY},1_{TZ})\) identify it with \((Tf,Tg,-Th)\): the first two squares each change two signs, and the last square changes none. This is exactly the functorial image using (3.1). Naturality of minus the identity follows from additivity. The underlying functor \(T\) is an equivalence by hypothesis.

For completeness this structure reflects distinguished triangles as well. If \((Tf,Tg,-Th)\) is distinguished, take three inverse rotations and cancel the coherent \(ST\) comparisons. The result is \((-f,-g,h)\), isomorphic to (2.1) by \((1_X,-1_Y,1_Z)\).

An explicit inverse structure on \(S\) has comparison

For the composite \(TS\), its comparison is \((T\epsilon_X)^{-1}(T\eta_X)^{-1}\). The counit is a translation-compatible natural isomorphism from this composite to the identity, because

The last equality is the triangular identity \(\epsilon_{TX}T\eta_X=1\). Consequently applying \(T\) to the \(S\)-image of a distinguished triangle produces a triangle isomorphic through \(\epsilon\) to the original one. Reflection just proved shows that this \(S\)-image is distinguished. The unit supplies the other inverse comparison by the other triangular identity. \(\square\)

4. An idempotent tower has an ordinary colimit

Suppose countable coproducts exist and \(p:X\to X\) satisfies \(p^2=p\). Put \(P=\bigoplus_{n\ge0}X\), with injections \(a_n\), and specify

Choose a distinguished triangle

This chosen \(Y\) is the homotopy colimit of the constant-object tower with transition \(p\).

Theorem 4.1. The maps in (4.2) determine \(q:X\to Y\) and \(r:Y\to X\) satisfying

The object \(Y\), with every structure map equal to \(q\), is the ordinary categorical colimit of that tower. In particular every idempotent in a triangulated category with countable coproducts splits.

Proof. Define a map \(t:P\to P\) by its finite columns

This is a legal coproduct map: each column is a finite sum, and the coproduct universal property assembles the columns. Idempotence gives

Thus \(td=1_P\). Theorem 2.2 applied to (4.2) gives \(h=0\) and makes \(v\) a split epimorphism.

Since \(d a_n(1-p)=a_n(1-p)\) and \(vd=0\), we have \(va_n=va_np\). Equation (4.1) now gives \(va_n=va_{n+1}\). Write their common value as \(q\); then \(q=qp\).

Define \(b:P\to X\) by \(ba_n=p\). Its columns show \(bd=0\), so exact \(\operatorname{Hom}(-,X)\) in (4.2) gives \(r:Y\to X\) with \(rv=b\). Hence \(rq=p\). The maps \(qrv\) and \(v\) agree on every \(a_n\), since \(qba_n=qp=q=va_n\). The epimorphism \(v\) cancels to give \(qr=1_Y\). This proves the splitting with its specified maps.

A cocone into \(W\) is a sequence \(u_n:X\to W\) with \(u_n=u_{n+1}p\). This relation implies \(u_n p=u_n\), and then \(u_n=u_{n+1}\). Thus it is one map \(u\) with \(up=u\). The map \(ur:Y\to W\) satisfies \((ur)q=up=u\). It is unique: if \(\phi q=u\), then \(\phi=\phi qr=ur\). These are exactly the ordinary colimit universal property and all its structure-map equations. No interchange of Hom with countable coproducts, compactness assumption, or general exactness of filtered colimits has been used. \(\square\)

Example 4.2. In vector spaces let \(X=U\oplus V\) and \(p(u,v)=(u,0)\). The colimit is \(U\), with \(q(u,v)=u\) and \(r(u)=(u,0)\). Formula (4.4) keeps the \(V\)-component in column \(n\) and subtracts the \(U\)-component in columns \(0,\ldots,n-1\). These finite columns verify (4.5) even when \(U,V\) have arbitrary dimension.

5. Abelian triangulated categories are semisimple

An abelian category is semisimple here if every short exact sequence splits. This definition does not impose finite length, finite dimension, or a specified set of simple objects. Use the actual kernel, image, quotient and biproduct constructions of Exact complexes, signs and lifting covers, Sections 1–2.

Theorem 5.1. If the underlying category of a triangulated category is abelian, it is semisimple. Every distinguished triangle is isomorphic to a finite direct sum of the three elementary types

The last type can equivalently be written \(SZ\to0\to Z\to TSZ\), with last arrow \(\epsilon_Z^{-1}\).

Proof. In every short exact sequence, its first map is a categorical monomorphism. Corollary 2.3 supplies a retraction, and abelian exactness identifies the complement with the stated cokernel. Explicitly, if \(i:A\to B\) has retraction \(k\), its cokernel \(c:B\to C\) restricts to an isomorphism from \(\ker k\) to \(C\): write \(b=ikb+(b-ikb)\), or use the corresponding biproduct identity; injectivity and surjectivity of that restriction follow from \(\ker c=\operatorname{im}i\). This proves semisimplicity.

Factor an arbitrary \(f:X\to Y\) through its image \(I\). The exact sequences with kernel \(K\) and cokernel \(Q\) split, giving identifications

The notation denotes biproduct component maps, so this formula is meaningful without underlying elements. The finite sum of the three triangles (5.1), for \(I,Q,K\) respectively, is the distinguished completion

The third elementary type is distinguished: rotating the identity triangle of \(K\) gives \(K\to0\to TK\xrightarrow{-1}TK\), and minus the identity on its third vertex changes that last map to \(1\). Compare (5.3) with the given distinguished completion of \(f\), fixing \(X,Y\). The two-isomorphism comparison theorem gives an invertible third map preserving both remaining arrows. This gives the entire stated decomposition, including its last arrow. \(\square\)

Only distinguished triangles have this assertion. An arbitrary sequence of three arrows can fail the zero-composite or exactness conditions; calling every such sequence a sum of (5.1) would be false.

6. Constructing all axioms from semisimplicity

Conversely let \(\mathcal C\) be a semisimple abelian category and let \(T\) be an additive equivalence. Declare a triangle distinguished if it is isomorphic to a finite sum of (5.1). We prove all four axioms, including the axiom for three prescribed completions of a composable pair.

Lemma 6.1 (an exactness characterization). A triangle \(A\xrightarrow fB\xrightarrow gC\xrightarrow hTA\) is distinguished under this declaration if and only if

as subobjects, and its consecutive composites, including \((Tf)h\), are zero.

Proof. The normal triangle (5.3) has these properties by its component maps. They are invariant under triangle isomorphisms. For the converse choose (5.2) for \(f\). The first equality makes \(g\) vanish on \(I\) and induce a monomorphism \(\bar g:Q\to C\). Equivalences preserve kernels: applying an inverse equivalence to a proposed kernel factorization proves its universal property, and hence \(\ker(Tf)=TK\). The other two equalities make \(h\) factor through \(TK\) and give a short exact sequence

Here \(\bar h\) is epic because its image is the whole \(TK\); its kernel is \(\bar g(Q)\) by the middle equality. Semisimplicity supplies \(s:TK\to C\) with \(\bar h s=1\). The map \([\bar g,s]:Q\oplus TK\to C\) is invertible: subtracting \(s\bar h\) factors uniquely through \(\bar g\), giving its inverse components. This identifies the triangle with (5.3), preserving \(g,h\). \(\square\)

TR1 and TR2. The declaration is closed under triangle isomorphism, contains each identity triangle, and completes every \(f\) through (5.2)–(5.3). This is TR1. An equivalence preserves the actual image and kernel comparisons and finite biproducts by their universal properties. Consequently applying \(T\) to (6.1) gives its translated exact positions. The rotated arrows \((g,h,-Tf)\) satisfy the three required equalities at \(C,TA,TB\); a minus sign has the same kernel and image as the original map. Lemma 6.1 makes the rotation distinguished. Applying the inverse equivalence gives the converse, with its inverse-rotation sign. This proves both directions of TR2.

TR3. Choose normal forms (5.3) for the source and target triangles. Write \(\alpha:K\oplus I\to K'\oplus I'\) and \(\beta:I\oplus Q\to I'\oplus Q'\) for the first two vertex maps. The square \(\beta f=f'\alpha\) forces exactly

Define the third map from \(Q\oplus TK\) to \(Q'\oplus TK'\) by

The equality \(\gamma j=j'\beta\) is the first diagonal component and \(\beta_{Q'I}=0\). The equality \(e'\gamma=(T\alpha)e\) is the second diagonal component and \(\alpha_{I'K}=0\). Thus all three squares commute. Conjugate through the chosen normal-form isomorphisms to complete any original square. No claim of uniqueness is needed.

A six-piece decomposition for TR4. For \(A\xrightarrow fB\xrightarrow gC\), let \(I=\operatorname{im}f\), \(J=\ker g\), and let \(L=I\cap J\), defined by the pullback of their inclusions in \(B\). Split \(I=L\oplus M\) and \(J=L\oplus N\). Their union inside \(B\) is \(L\oplus M\oplus N\): the only intersection of \(I,J\) is \(L\), so the sum map is monic after these complements. Split that inclusion and write

The restriction of \(g\) to \(M\oplus R\) is monic, since its kernel lies in \(J=L\oplus N\). Its image splits in \(C\), giving \(C\cong M\oplus R\oplus V\) with \(g\) the identity on these two image summands and zero on \(L,N\). Split the epimorphism \(A\to I\) and obtain \(A\cong K\oplus L\oplus M\), with \(f\) the identity on \(L,M\) and zero on \(K\). The pair is therefore a finite direct sum of the six pairs

The summands correspond respectively to \(K,N,V,L,R,M\). This construction uses only kernels, pullbacks, images, and the given splitting of short exact sequences; it does not assume that objects have finite length.

TR4 for the normal completions. Choose the normal cone objects \(Q_f,Q_{gf},Q_g\), inclusions \(j_f,j_{gf},j_g\), and connecting maps \(e_f,e_{gf},e_g\) from (5.3). The required octahedral triangle is

and its four squares are

For the six summands (6.6), specify these data by the following table. A displayed \(1\) is the identity of the indicated nonzero object, and every other indicated map is zero.

We check all four equations and the prescribed last arrow. In the first row, \(e_f=e_{gf}=1_{TU}\), and all \(j\)'s and \(e_g\) are zero; only the second equation is nonzero and it reads \(1=1\). In the second row, \(j_f=1_U,e_g=1_{TU}\), with all other \(j\)'s and \(e\)'s zero. All four squares are zero and \(w=(Tj_f)e_g=1_{TU}\). In the third row, \(j_{gf}=j_g=1_U\), all connecting maps vanish, and only the third equation is nonzero: \(1=1\).

In the fourth row, \(e_{gf}=e_g=1_{TU}\), \(Tf=1_{TU}\), and every \(j\) is zero. Only the fourth equation is nonzero and says \(1=1\). In the fifth row, \(j_f=j_{gf}=1_U,g=1_U\), all connecting maps and \(j_g\) vanish, and only the first equation is nonzero and says \(1=1\). In the sixth row all cone maps and all equations are zero. In every row (6.7) is one of the elementary types (5.1), or the zero triangle. In particular the second row retains the positive identity connecting arrow of the third elementary type. Taking the finite sum proves (6.7)–(6.8) for the pair before normal-form conjugation; conjugation through the decomposition gives them for the original \(A,B,C,f,g\).

TR4 for prescribed completions. This last step is needed when the three input triangles are already given. TR3 and the exact Hom comparison theorem, which uses only TR1–TR3, give isomorphisms

fixing the first two vertices of each completion. Transport \(u,v\) to \(c_{gf}u c_f^{-1},c_gv c_{gf}^{-1}\). Transport \(w\) to \((Tc_f)w c_g^{-1}\). Since \(c_f j_f=\widetilde j_f\) and \(\widetilde e_g c_g=e_g\), this last map is exactly \((T\widetilde j_f)\widetilde e_g\). The remaining square identities follow by substituting the same two comparison equations for each of the three cones. Isomorphism closure makes the transported triangle distinguished. This verifies the full prescribed-completion form of TR4 and finishes the construction of a triangulated category. \(\square\)

Example 6.2. Vector spaces with \(T=1\) carry this structure. A normal triangle for a linear map \(f\) has third vertex \(\operatorname{coker}f\oplus\ker f\), with its inclusion and projection as in (5.3). A short exact sequence of vector spaces becomes a distinguished triangle with zero connecting map after choosing a splitting. The structure accommodates nonzero connecting maps as well: \(U\to0\to U\xrightarrow1U\) is distinguished. Semisimplicity therefore does not say that every last arrow vanishes.

7. Four graded exercises with complete solutions

Exercise 1 (basic). In vector spaces with \(T=1\), take \(f:k^3\to k^2\), \(f(a,b,c)=(a,b)\). Give the normal triangle, with every arrow, and decide whether \(f\) is a categorical monomorphism.

Solution. The kernel is the third-coordinate copy \(K=k\), the image is \(I=k^2\), and the cokernel is zero. The triangle is \(k^3\xrightarrow f k^2\xrightarrow0 k\xrightarrow h k^3\), with \(h(c)=(0,0,c)\). At \(k^2\), the image of \(f\) is the whole kernel of zero. At \(k\), the image of zero and the kernel of \(h\) are both zero. At \(k^3\), the image of \(h\) equals the kernel of \(f\). Thus Lemma 6.1 verifies that it is distinguished. The two maps \(k\to k^3\) given by zero and by \(c\mapsto(0,0,c)\) are distinct and have equal composites with \(f\). Therefore \(f\) is not monic. It is epic and has section \((a,b)\mapsto(a,b,0)\), in agreement with Corollary 2.3.

Exercise 2 (intermediate). Let \(p:k^2\to k^2\) be \(p(a,b)=(a,0)\). Compute \(td a_2\) by (4.4), and prove directly that the ordinary colimit of its tower is \(k\).

Solution. The second and third finite columns are \(t a_2=a_2(1-p)-a_0p-a_1p\) and \(t a_3=a_3(1-p)-a_0p-a_1p-a_2p\). Their difference after multiplying the latter by \(p\) is

For any target vector space \(W\), a compatible family \(u_n:k^2\to W\) satisfies \(u_n=u_{n+1}p\). Hence all \(u_n\) coincide and kill the second coordinate. There is a unique map \(\phi:k\to W\) with \(\phi(a)=u_n(a,0)\); its composite with \(q(a,b)=a\) equals each \(u_n\). Uniqueness follows by evaluating at \((a,0)\). This is the colimit property, including all its maps. The section \(r(a)=(a,0)\) gives \(rq=p,qr=1\), so it is the idempotent splitting and agrees with the chosen telescope through the unique colimit isomorphism.

Exercise 3 (advanced). For \(f(a,b)=(a,0)\) on \(k^2\), use the source and target normal triangle of this same map. Let

Verify the first square and give a third comparison. Then determine all possible third comparisons in these coordinates, with \(T=1\).

Solution. Both \(\beta f\) and \(f\alpha\) send \((a,b)\) to \((a,0)\). The kernel coordinate is the second coordinate of the source, the image coordinate is the first, and the cokernel coordinate is the second coordinate of the target. Thus \(Q_f=k\oplus k\), \(j(a,b)=(b,0)\), and \(e(q,z)=(0,z)\). Formula (6.4) gives \(\gamma(q,z)=(5q,4z)\). Then \(\gamma j(a,b)=(5b,0)=j\beta(a,b)\) and \(e\gamma(q,z)=(0,4z)=\alpha e(q,z)\). For a general matrix \(\begin{psmallmatrix}x&y\\z&t\end{psmallmatrix}\), the first equality forces \(x=5,z=0\), and the second forces \(t=4\). It imposes no condition on \(y\). Hence all comparisons are \(\gamma_y(q,z)=(5q+yz,4z)\), for \(y\in k\), interpreting the indicated integers in \(k\). This verifies actual existence and shows the remaining nonuniqueness. No invertibility was required of \(\alpha,\beta,\gamma\).

Exercise 4 (advanced). In vector spaces with \(T=1\), set \(A=k^2,B=k^3,C=k^2\),

Find the six interval summands and all three normal cones. Compute \(u,v,w\) and verify the octahedron with these original coordinates.

Solution. The \(a\)-coordinate is the interval \([1,2]\): it maps identically to the \(x\)-coordinate and then to zero. The \(b\)-coordinate is \([1,2,3]\), passing identically through \(y\) to the first coordinate of \(C\). The \(z\)-coordinate is \([2,3]\), passing identically to the second coordinate of \(C\). The other three interval objects are zero. Thus \(Q_f=k\), with \(j_f(x,y,z)=z\) and \(e_f=0\); \(Q_{gf}=k\oplus k\), with first coordinate the second-coordinate cokernel in \(C\), second coordinate the \(a\)-kernel in \(A\); and \(Q_g=k\), the \(x\)-kernel in \(B\). Their maps are

The first equation of (6.8) is \(u j_f(x,y,z)=(z,0)=j_{gf}g(x,y,z)\). The second is \(e_{gf}u=0=e_f\). The third is \(v j_{gf}=0=j_g\). The fourth is \(e_gv(s,a)=(a,0,0)=f e_{gf}(s,a)\). Finally \(j_f e_g=0\), so the prescribed last arrow is indeed \(w=0\). The octahedral triangle is \(k\to k\oplus k\to k\to0\), with its canonical first injection and second projection. Theorem 2.2, or the defining sum of elementary triangles, proves it distinguished. These calculations check every arrow on the original coordinates, rather than just recording the isomorphism classes of its cones.

8. Freely accessible references

The proofs above are complete within the linked course prerequisites. For a freely accessible parallel treatment of the zero-cone criterion, split triangles, categorical kernels and the transport of octahedra, see the Stacks Project, Derived Categories, Section 13.4, especially Lemmas 13.4.9, 13.4.11, 13.4.12 and 13.4.15. The countable idempotent argument above also proves the actual ordinary colimit identification; the mere statement that a category is idempotent complete would not supply that identification.