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For a cohomological spectral sequence, $d_r:E_r^{p,q}\to E_r^{p+r,q-r+1}$ and the next page is its cohomology. In the bounded setting, convergence of a spectral sequence to $H^{\ast }$ means that each $H^n$ has a finite decreasing, exhaustive and separated filtration withHere $E_{\infty }^{p,q}$ is the eventual stable value. This determines the associated graded module of $H^n$, rather than automatically a canonical direct-sum decomposition of $H^n$. For unbounded filtrations additional completeness/convergence conditions are necessary; boundedness removes those issues here.

The bounded filtered-complex convergence theorem states the following. If $F^{p}C$ is a decreasing filtration by cochain subcomplexes, preserved by the differential, finite in each cochain degree (uniform bounds $F^{a}C=C$, $F^{b}C=0$ are sufficient), then there is a spectral sequenceThe abutment filtration is $F^p{H}^n(C)=\mathrm{im}[H^n(F^{p}C)\to H^n(C)]$. Its finite length ensures stabilization and the displayed limiting-page identification. The filtered cochain complex need not itself be bounded in cochain degree. In the degreewise finite version, the bounds in degrees $n-1,n,n+1$ suffice to stabilize the terms of total degree $n$.

For a double cochain complex $D^{p,q}$ bounded in both indices, form the total cochain complex ${\mathrm{Tot}}^{n}D={⨁}_{p+q=n}{D}^{p,q}$ with anticommuting differentials $d_h,d_v$ and total differential $d_h+d_v$. If the original differentials commute, inserting the usual sign in one of them gives this convention. Filtering by the first index gives $E_1^{p,q}=H_v^q(D^{p,\ast })$ and then $E_2^{p,q}=H_h^p(H_v^{q}D)$. Filtering by the second index gives the spectral sequence taking horizontal cohomology first and vertical cohomology next. Both filtrations are finite, so both converge to $H^{\ast }(\mathrm{Tot}D)$. This is the two spectral sequences of a bounded double complex construction.

The printed left/left tensor expression needs a handedness repair. Over an arbitrary ring, a tensor product of modules over $R$ pairs a right module with a left module. To retain the order of the printed formula, take $P$ to be a bounded cochain complex of projective right $R$-modules and $M$ a left $R$-module. Alternatively, keep $P$ left, take $M$ right, and write $M{\otimes }_{R}P$ and ${\mathrm{Tor}}_{-p}^R(M,H^{q}P)$. For a commutative ring no repair is needed. We prove the first, correctly typed formulation.

Finite projective dimension gives a finite projective resolution $Q_{\bullet }\to M$ by left modules. Form $D^{p,q}=P^q{\otimes }_R{Q}_{-p}$ for $-d\le p\le 0$. On $x\otimes y$ take $d_v=d_P\otimes 1$ and $d_h=(-1{)}^{q}1\otimes d_Q$. These anticommute, and the double cochain complex is bounded in both directions. Projective modules are flat modules, so taking vertical cohomology first givesThe first identification uses flatness of $Q_{-p}$; the second is the Tor functor computed by resolving its left-module argument $M$. Balancedness of the Tor functor allows either correctly sided projective resolution to compute it, by the double-resolution argument.

Taking horizontal cohomology first instead uses flatness of $P^q$. The resolution $Q_{\bullet }$ then leaves only $P^q{\otimes }_{R}M$ in column $p=0$. Its remaining differential is $d_P\otimes 1$, so the augmentation $\mathrm{Tot}D\to P{\otimes }_{R}M$ is a quasi-isomorphism. Apply the convergence theorem to this underlying abelian-group double complex. The two spectral sequences therefore prove the Künneth spectral sequence:These are abelian groups in general; an extra module structure requires appropriate bimodule hypotheses. The index $p$ is nonpositive, so $-p$ is the nonnegative homological index of the Tor functor.

Finally write the cycle modules as $Z^q=\ker d^q$ and the boundary modules as $B^q=\mathrm{im}{d}^{q-1}$. If every boundary module $B^q$ is projective, the short exact sequence $0\to Z^q\to P^q\to B^{q+1}\to 0$ splits. Thus $Z^q$ is projective. The short exact sequence $0\to B^q\to Z^q\to H^{q}P\to 0$ is now a length-one projective resolution, so ${\mathrm{Tor}}_i^R(H^{q}P,M)=0$ for $i\ge 2$. The $E_2$ page occupies only columns $p=-1,0$. Every $d_r$ for $r\ge 2$ goes $r$ columns to the right, hence has zero source or zero target. ThereforeThis two-column degeneration of a Künneth spectral sequence gives the edge short exact sequencesDegeneration itself does not supply a canonical splitting. Projective boundaries also do not force $H^{q}P$ to be projective: a complex with differential multiplication by $2$ over $\mathbb{Z}$ has projective boundary but a $\mathbb{Z}/2\mathbb{Z}$ cohomology group.

Schur's lemma says that a nonzero intertwiner between irreducible representations is an isomorphism; over $\mathbb{C}$, every endomorphism of a finite-dimensional irreducible representation is scalar. To prove the Schur lemma, let $T:V\to W$ be a intertwining operator between irreducible representations. Its kernel and image of a linear map are invariant. If $T\ne 0$, irreducibility forces $\ker T=0$ and $\mathrm{im}T=W$. This proves the first assertion over any field. In particular the endomorphisms of an irreducible representation form a division ring.

When $V$ is finite-dimensional over an algebraically closed field, an endomorphism $T$ has an eigenvalue $\lambda$. The endomorphism $T-\lambda I$ has nonzero kernel. By the first assertion it must be zero, so $T=\lambda I$. Consequently, for complex finite-dimensional irreducible representations, the space of intertwining operators has dimension zero for nonisomorphic representations and dimension one for isomorphic representations.

Every finite-dimensional representation of a complex semisimple Lie algebra is completely reducible. We prove the Weyl complete reducibility theorem using the allowed Casimir operator facts, without assuming a splitting in advance. For dual bases $x_i,x^i$ with respect to the Killing form, the Casimir elementis central in the universal enveloping algebra. Hence its Casimir operator $C_M={\sum }_i{\rho }_M(x_i){\rho }_M(x^i)$ commutes with the action on every Lie algebra representation $M$ and is compatible with subrepresentations, quotient representations, and intertwining operators. On the trivial Lie algebra representation it acts by zero. On every nontrivial finite-dimensional Irreducible Lie algebra representation it acts by a nonzero scalar.

For clarity, the last fact can be expressed by the Casimir eigenvalue formula: on an irreducible with dominant integral weight $\lambda$, the scalar is $(\lambda ,\lambda +2\rho )$, with the inner product induced by the Killing form and $\rho$ the half-sum of positive roots. On the real span of the weights this inner product is positive definite, and the scalar is positive for $\lambda \ne 0$. For a semisimple Lie algebra with several simple factors the scalars add, so a nontrivial representation still gives a nonzero scalar. These are properties of the Casimir operator being used here.

First establish Casimir splitting of a trivial quotient. Supposeis a short exact sequence of finite-dimensional Lie algebra representations, with trivial quotient. The generalized eigenspaces of $C_E$ are invariant, soEach $E_c$ with $c\ne 0$ maps to zero under $q$: applying a sufficiently large power of $C_E-cI$ and using $C_{\mathbb{C}}=0$ gives $(-c{)}^{r}q(e)=0$. Thus $q(E_0)=\mathbb{C}$.

Take a composition series of a module of $E_0$. The Casimir operator is nilpotent on $E_0$, so its scalar on every irreducible composition factor is zero. The stated Casimir operator property makes every such factor trivial. In a basis adapted to the composition series of a module, the image of $\mathfrak{g}$ on $E_0$ therefore consists of strictly upper triangular matrices, so that image is solvable. The allowed fact that a semisimple Lie algebra acts trivially on every one-dimensional representation implies that $\mathfrak{g}$ is a perfect Lie algebra: otherwise a nonzero linear functional on $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ would define a nontrivial one-dimensional Lie algebra representation. Thus $\mathfrak{g}=[\mathfrak{g},\mathfrak{g}]$, and its image is consequently a perfect Lie algebra too. A perfect Lie algebra that is also a solvable Lie algebra is zero, since its derived series of a Lie algebra is constant until it vanishes. Hence $\mathfrak{g}$ acts trivially on $E_0$. Choose $e\in E_0$ with $q(e)=1$. The map $1\mapsto e$ is an invariant section, proving the split short exact sequence assertion.

Now let $W\subseteq V$ be any nonzero subrepresentation. On the Hom representation $\mathrm{Hom}(V,W)$ the action isConsider the invariant subspaceRestriction produces a short exact sequenceSurjectivity follows by extending $I_W$ to a linear map on $V$. The quotient is trivial, because $I_W$ commutes with the action on $W$. The preceding Casimir splitting of a trivial quotient yields an invariant $p:V\to W$ with $p{|}_W=I_W$. Thusand $\ker p$ is invariant. The zero subrepresentation also has a complement. Repeatedly splitting off an irreducible subrepresentation now gives a direct sum of irreducibles, proving the Weyl complete reducibility theorem.

A derivation of a Lie algebra is a linear map $D:\mathfrak{g}\to \mathfrak{g}$ satisfying the Leibniz ruleThe space $\mathrm{Der}(\mathfrak{g})$ is a vector subspace of $\mathrm{End}(\mathfrak{g})$. Equip it with the commutator $[D,E]=DE-ED$. Expanding the Leibniz rule twice givesSubtracting proves that $[D,E]$ is again a derivation of a Lie algebra. Antisymmetry and the Jacobi identity hold for the commutator in every associative endomorphism algebra, so this defines the derivation Lie algebra.

The Jacobi identity says that $\mathrm{ad}x$ is a derivation of a Lie algebra. Moreover, for $D\in \mathrm{Der}(\mathfrak{g})$,soThis makes $\mathrm{ad}\mathfrak{g}$ an ideal of a Lie algebra in $\mathrm{Der}(\mathfrak{g})$.

Finally suppose $\mathfrak{g}$ is semisimple. Let $\mathfrak{g}$ act on $\mathrm{Der}(\mathfrak{g})$ by $x\cdot D=[\mathrm{ad}x,D]$. The Weyl complete reducibility theorem supplies an invariant complement $M$ to $\mathrm{ad}\mathfrak{g}$. For $D\in M$, invariance gives $[\mathrm{ad}x,D]\in M$, while the ideal identity above gives $[\mathrm{ad}x,D]\in \mathrm{ad}\mathfrak{g}$. Their intersection is zero, so $\mathrm{ad}(Dx)=0$ for every $x$. The center of a Lie algebra of a semisimple Lie algebra is zero, hence $Dx=0$ for every $x$, and $M=0$. ThereforeEvery derivation of a Lie algebra is inner, and the element giving it is unique because the center of a Lie algebra vanishes.

First account for the unlabelled coefficient-sequence construction. The two short exact sequences of abelian groups areThe singular chain groups of $X$ are free abelian. Applying $\mathrm{Hom}(C_q(X),-)$ therefore preserves these exact sequences, degree by degree, giving short exact sequences of cochain complexes. The associated long exact sequence from a coefficient sequence gives the displayed maps in cohomology; the connecting maps are the integral Bockstein homomorphism $\hat{\beta }$ and the modulo-$n$ Bockstein homomorphism $\beta$. The first omitted map is multiplication by $n$, and the second is induced by $j$.

For the requested example, attach an $(i+1)$-cell to $S^i$ using a map of degree $n$. The resulting Moore space $X=M(\mathbb{Z}/n,i)$ has positive-degree cellular chain complexin degrees $i+1,i$. This construction also works for $i=1$, using the degree-$n$ map of the circle. In cellular cohomology with coefficients $\mathbb{Z}/n$, the differential is zero, so both $H^i$ and $H^{i+1}$ are $\mathbb{Z}/n$.

Lift the cochain taking value $1$ on the $i$-cell to a cochain with coefficients $\mathbb{Z}/n^2$. Its coboundary takes value $n$ on the $(i+1)$-cell, which is $j(1)$. The definition of the connecting homomorphism therefore sends the degree-$i$ generator to the degree-$(i+1)$ generator. HenceIt is nonzero for every $i\ge 1$ and $n\ge 2$, including composite $n$. This is the Bockstein on a cyclic Moore space.

Put $R_j=S/I_j$ and $R_0=S$. The hypothesis says multiplication by $F_j$ on the graded ring $R_{j-1}$ is injective for $j=2,3$. Checking homogeneous elements suffices: an arbitrary annihilated element splits into homogeneous components, each annihilated separately because $F_j$ and $I_{j-1}$ are homogeneous. Also $F_1\ne 0$ because it has degree $d_1>0$, and $S$ is an integral domain. Thus $F_1,F_2,F_3$ is a regular sequence, and the suggested short exact sequences becomeSheafification and twisting preserve exactness. With $X_0={\mathbb{P}}_{\mathbb{C}}^4$, the resulting sequences, interpreted on the ambient projective space via the closed inclusions, areCohomology under a closed immersion identifies the displayed sheaf cohomology with that on $X_j$. For every integer $m$, cohomology of twisting sheaves on projective space gives $H^q(X_0,\mathcal{O}(m))=0$ for $1\le q\le 3$. Inducting through the long exact sequence in sheaf cohomology givesIndeed the two adjacent groups have degrees $q$ and $q+1$ on $X_{j-1}$, both in its vanishing range. This is intermediate cohomology vanishing for a projective complete intersection.

We simultaneously prove $H^0(X_j,{\mathcal{O}}_{X_j}(m))=0$ for $m<0$ and $H^0(X_j,{\mathcal{O}}_{X_j})=\mathbb{C}$. Both hold on $X_0$ by the same projective-space formula. For $j\le 3$, the preceding stage has $H^1(X_{j-1},\mathcal{O}(m-d_j))=0$. If $m<0$, both degree-zero groups on that stage vanish, so the new one vanishes. If $m=0$, the left degree-zero group vanishes because $-d_j<0$, while the middle group is $\mathbb{C}$. Therefore the natural restriction of constant functions is an isomorphism at every stage:These are isomorphisms of $\mathbb{C}$-algebras, not just vector spaces. The global regular functions on a positive-dimensional projective complete intersection are constants even if the complete intersection is singular or nonreduced; smoothness and reducedness were not assumed.

Use integer coefficients. The mapping-torus homology from Mayer–Vietoris works for every continuous map $f:X\to X$, without assuming that the mapping torus is a fiber bundle. Project to the time circle and choose two open arcs, one containing the seam and one away from it. The inverse image $U$ of the seam arc is a mapping cylinder neighborhood with a deformation retraction to its target copy of $X$; the inverse image $V$ of the other arc is a product with a deformation retraction to $X$. Their intersection has two components, each with a deformation retraction to $X$. Label the upper component first. With the convention that the second inclusion in the Mayer–Vietoris sequence has a minus sign, its homology map isIndeed, the upper component includes into the seam neighborhood by the identity, while retracting the lower component across the seam applies $f$. Put $u=a+b$ and $v=b$, and replace target coordinates $(p,q)$ by $(p+q,-q)$. The map becomes $((f_{\ast }-\mathrm{id})v,u)$. Negating $v$ gives a direct sum of $\mathrm{id}-f_{\ast }$ and an isomorphism. Canceling the latter summands in the long exact sequence yieldsIn particular, there is a short exact sequenceNo splitting of this short exact sequence is assumed at this stage.

For the torus, take the coordinate loops $c_1(t)=[(t,0)]$ and $c_2(t)=[(0,t)]$ as the basis of $H_1$. Their lifts to ${\mathbb{R}}^2$ have endpoint displacements $e_1,e_2$. Applying $A$ changes those displacements to its columns, soThis also follows from the identification of the first homology group with the abelianization of the fundamental group. The Künneth theorem gives $H_2(X)=\mathbb{Z}$, with generator the product orientation class. If $\alpha ,\beta$ are the dual degree-one cohomology classes, then $\alpha ⌣\beta$ evaluates to one on this generator, andThus the degree-two homology map is multiplication by $\det A$, as in homology of an integer torus endomorphism. The claim that the torus map is a homeomorphism requires $A$ to be a unimodular matrix, equivalently an integer inverse or $\det A=\pm 1$; nonsingularity over $\mathbb{R}$ alone would only give a covering map of finite degree. The particular matrix here is unimodular.

For this matrix, $\det A=1$ andIts kernel is zero and its Smith normal form is $\mathrm{diag}(1,2)$, so its cokernel is $\mathbb{Z}/2$. The maps $\mathrm{id}-f_{\ast }$ on $H_0$ and $H_2$ are zero. Substituting into the short exact sequences, and splitting the degree-one extension because its quotient $\mathbb{Z}$ is free, gives

The roots-of-unity criterion for a splitting field for finite group representations says that the assumed $m$th roots of unity make $k$ a splitting field for $G$. Fix a multiplicative identification of the group ${\mu }_m(k)$ with the complex $m$th roots of unity. For a p-regular element $g$, its order divides $m$; the operator $\rho (g)$ is diagonalizable because $X^{|g|}-1$ has distinct roots in $k$. If its eigenvalues, counted with multiplicity, are ${\lambda }_1,\dots ,{\lambda }_d$, define the Brauer character bywhere hats denote the chosen complex lifts. This defines a class function on the p-regular elements, not on arbitrary elements of $G$.

A short exact sequence can be represented by block triangular matrices, so the eigenvalue multiset is the union of those on its submodule and quotient. Thus Brauer characters are additive on short exact sequences. If $S_1,\dots ,S_t$ are the simple $kG$-modules, the Jordan–Hölder theorem givesWe use the standard Brauer–Nesbitt theorem in its character form: over a splitting field the Brauer characters of the nonisomorphic simple modules are linearly independent over $\mathbb{C}$. ThereforeBoth the modular splitting-field criterion and this independence theorem are the representation-theoretic results used here.

We prove $(iii)\Rightarrow (ii)$. Consider one nonsplit short exact sequencewith $U,W$ simple. Lift the canonical surjection $P(W)\to W$ from the projective cover through $q$, obtaining $f:P(W)\to V$. Its image maps onto $W$. The intersection $\mathrm{im}f\cap U$ is either zero or $U$, since $U$ is simple. If it were zero, $q$ would identify $\mathrm{im}f$ with $W$, splitting the extension. Hence $U\subset \mathrm{im}f$, and surjectivity onto $W$ then implies $\mathrm{im}f=V$.

Therefore both $U$ and $W$ occur as Jordan–Hölder factors of the indecomposable projective $P(W)$. Replacing every edge in an Ext chain by this observation proves (ii). Together with the preceding two implications, this provesThe repeated word “projective” in the printed version of (ii) has no mathematical effect.

Use the standard complex K-theory Thom class convention. For a rank-$r$ complex vector bundle, its K-theory Euler class is the pullback of the K-theory Thom class along the zero section, namelyHere ${\Lambda }^j$ denotes an exterior power, and a Hermitian metric identifies the dual bundle with the conjugate vector bundle. Choosing ${\lambda }_{-1}(E)$ instead changes the K-theory Euler class by the unit $(-1{)}^r[\det E]$, so its kernel and cokernel computations are unchanged.

For the disk bundle $D(E)$ and sphere bundle $S(E)$, the K-theory Thom class identifies $K^i(D(E),S(E))$ with $K^i(X)$. The exact sequence of this pair becomes the K-theory Gysin sequence of a sphere bundle:The groups are two-periodic by Bott periodicity.

View Real projective space as the circle bundle of ${\gamma }^{\otimes 2}$ over Complex projective space, where $\gamma \to {\mathbb{CP}}^n$ is the tautological bundle. Indeed, $S^{2n+1}/\{\pm 1\}={\mathbb{RP}}^{2n+1}$ maps to $S({\gamma }^{\otimes 2})$ by $[v]\mapsto v\otimes v$; in each fiber squaring identifies exactly $v$ and $-v$.

Use Complex K-theory of complex projective space in the formThe K-theory Euler class of ${\gamma }^{\otimes 2}$ is $1-[\gamma {]}^{-2}$. Multiplying it by the unit $-[\gamma {]}^2$ gives $1-[\gamma {]}^2=2t-t^2$. The K-theory Gysin sequence of a sphere bundle therefore givesUnder the projection $q:{\mathbb{RP}}^{2n+1}\to {\mathbb{CP}}^n$, the pullback vector bundle $q^{\ast }\gamma$ is the complexification of a real vector bundle $L$ of the tautological bundle of Real projective space: the fiber map sends $[v,z]$ to $zv$. Thus $t$ pulls back to $1-[L]$. Put $\mu =[L]-1$. The relation $L^{\otimes 2}\cong 1$ gives ${\mu }^2=-2\mu$, and $t^{n+1}=0$ gives $2^n\mu =0$. HenceFor completeness, these relations give precisely the abelian group $\mathbb{Z}\oplus \mathbb{Z}/2^n$. In the basis $t,\dots ,t^n$, the image of multiplication by $2t-t^2$ is generated by $2t-t^2,2t^2-t^3,\dots ,2t^n$; its relation matrix is triangular with determinant $2^n$, and the relations express every $t^j$ as $2^{j-1}t$. Thus there is exactly one cyclic torsion subgroup of order $2^n$.

If $n\ge 1$ and $a=a_0+a_{1}t+\cdots +a_n{t}^n$ is in the kernel, successive coefficients of $(2t-t^2)a$ give $2a_0=0$, $2a_1-a_0=0$, and so on, up to $2a_{n-1}-a_{n-2}=0$. Hence $a_0=\cdots =a_{n-1}=0$, with $a_n$ arbitrary. For $n=0$ the multiplier is zero on $R=\mathbb{Z}$. In all cases,

Finally let $E=L^{\oplus k}$ with $k\ge 1$, over $B={\mathbb{RP}}^{2n+1}$. Since $L^{\ast }\cong L$, multiplicativity of the K-theory Euler class givesOn $K^0(B)=\mathbb{Z}\oplus (\mathbb{Z}/2^n)\mu$, the image of multiplication by this K-theory Euler class is the subgroup generated by $2^{k-1}\mu$: multiplication by an integer produces every such multiple, while multiplication by $\mu$ only adds multiples of $2^k\mu$. ThereforeMultiplication by $\mu$ is zero on $K^{-1}(B)\cong \mathbb{Z}$, since $2^n\mu =0$ and this abelian group has no nonzero torsion elements. Thus the K-theory Gysin sequence of a sphere bundle gives a short exact sequenceIt splits as an exact sequence of abelian groups because its quotient is a free abelian group. ConsequentlyThe convention $\mathbb{Z}/1=0$ covers $n=0$ and $k=1$. For example, when $k=1$ the sphere bundle is $(S^{2n+1}\times S^1)/\{(v,z)\sim (-v,-z)\}$, a mapping torus of the antipodal map; its Topological K-theory has no nonzero torsion elements, agreeing with the formula. If $k=0$ is allowed, the rank-zero vector bundle has an empty unit sphere bundle, so $K^0(S(E))=0$ instead.

The tangent-normal short exact sequencegives $\det (TX{|}_Y)\cong \det (TY)\otimes N_{Y/X}$. Dualizing and using the canonical bundle,Since $N_{Y/X}\cong \mathcal{O}(Y){|}_Y$, rearrangement proves the adjunction formula

Suppose $X=A\cup B$ and that the interiors of $A$ and $B$ cover $X$. The Mayer–Vietoris theorem gives the long exact sequence in homologywhere all four displayed maps except the connecting homomorphism $\partial$ are induced by the relevant inclusion maps.

To prove it, let $C_{\ast }^{A,B}(X)$ be the chain complex generated by those singular simplices whose images lie wholly in $A$ or wholly in $B$. The sequenceis a short exact sequence of chain complexes. Repeated barycentric subdivision makes every singular simplex small enough to lie in $A$ or $B$, and the subdivision operator is chain homotopic to the identity. The inclusion $C_{\ast }^{A,B}(X)\hookrightarrow C_{\ast }(X)$ therefore induces an isomorphism on every homology group. The long exact homology sequence of a short exact sequence of chain complexes now gives the displayed sequence. Explicitly, if an $n$-cycle $z=a+b$ with $a\in C_n(A)$ and $b\in C_n(B)$, then $\partial a=-\partial b$ lies in $C_{n-1}(A\cap B)$ and represents the connecting class $\partial [z]$.

Let $T$ be the surface with boundary obtained by deleting the interiors of the $n$ disks. By excision,so $H_2(S,T)\cong {\mathbb{Z}}^n$ and the other relative groups vanish. In the long exact sequence in relative homology, the mapsends an orientation class to $(1,\dots ,1)$, up to an overall choice of sign. It is therefore injective and has cokernel ${\mathbb{Z}}^{n-1}$. Exactness gives $H_2(T)=0$ and a short exact sequenceThis sequence splits because its right-hand group is a free abelian group. Since $T$ is connected, we obtain

The short exact sequence of $G$-modules induces the long exact sequence in group cohomology

Regard the given short exact sequence as a degreewise short exact sequence of chain complexes concentrated in degree $n$. The inverse functor in the Dold–Kan correspondence is exact, so it produces a degreewise short exact sequence of simplicial abelian groupsThe last map is degreewise surjective and hence a Kan fibration. Its strict fiber is $K(A_1,n)$, and a strict fiber of a fibration computes the homotopy fiber. This proves the asserted homotopy fiber sequence of pointed Kan complexes.

Write the nonsplit short exact sequenceFor an endomorphism $g:M\to M$, the composite $qgj:L\to N$ vanishes because ${\mathrm{Hom}}_R(L,N)=0$. Hence $g$ restricts to an endomorphism $f$ of $L$ and induces an endomorphism $h$ of $N$, giving a commutative diagram of short exact sequences.

Because $L$ and $N$ are brick modules, each of $f,h$ is either zero or an isomorphism. If both are isomorphisms, the short five lemma makes $g$ an isomorphism. If both vanish, $g$ factors successively through $N$ and through $L$, hence through a map $N\to L$; this map is zero, so $g=0$.

The mixed cases would split the sequence. If $f$ is invertible and $h=0$, then $qg=0$, so $g=j\alpha$ for some $\alpha :M\to L$; the identity $\alpha j=f$ makes $f^{-1}\alpha$ a retraction of $j$. If $f=0$ and $h$ is invertible, then $gj=0$, so $g=\beta q$; the identity $q\beta =h$ makes $\beta h^{-1}$ a section of $q$. Both contradict nonsplitting. Thus every endomorphism of $M$ is zero or invertible, and $M$ is a brick.