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An opmonoidal functor $F:(\mathcal{C},\otimes ,I)\to (\mathcal{D},\otimes ,J)$ consists of a functor, a natural transformationand a morphism $F_0:FI\to J$. These maps need not be invertible. If $\alpha ,\lambda ,\rho$ are the respective associators and unitors, its axioms areThe first equation has domain $F((X\otimes Y)\otimes Z)$ and codomain $FX\otimes (FY\otimes FZ)$, which fixes the direction of every arrow. This is also called a colax monoidal functor.

An opmonoidal natural transformation $\tau :F\Rightarrow G$ between opmonoidal functors is a natural transformation satisfyingIt respects both the tensor and unit comparison maps. These are commutative diagrams and a unit commutative diagram; the arrow directions are opposite to those for a lax monoidal functor.

Because the direction of a lax monoidal functor compares the target tensor with the image of the source tensor, the given maps ${\phi }_{X,Y}:X\bullet Y\to X\diamond Y$ have the correct direction. They need not be invertible unless a strong monoidal functor is intended.

Their associativity condition is the following commutative diagram, with the associators of the two structures labelled explicitly:The two unit conditions, since ${\phi }_0=1_I$, are the unit commutative diagramsTogether with the assumed naturality, these are exactly the axioms for the requested monoidal structure. There are no further coherence conditions.

The faithful strict monoidal functor $U$ allows all three diagrams to be checked after applying it. Both tensor structures then have the same underlying tensor and constraints in $\mathcal{V}$. In particular, the unit conditions are equivalent toThe associativity square becomes the displayed diagram with every tensor replaced by $\otimes$, both associators by ${\alpha }^{\mathcal{V}}$, and every $\phi$ by $U\phi$. Faithfulness reflects equality of the two resulting composites. One should not identify ${\phi }_{I,X}$ itself with an identity in $\mathcal{C}$: its domain and codomain can be distinct objects with the same image under $U$. Intrinsically it is $({\lambda }_X^{\diamond }{)}^{-1}{\lambda }_X^{\bullet }$, and similarly on the right.

The Snake lemma associates to a commutative diagram with exact rows an exact sequenceFor $K=L_1\cup L_2$, apply it degree by degree to the short exact sequence of chain complexesThe resulting connecting maps splice the kernels modulo boundaries into the Mayer--Vietoris sequence

Now let $K$ satisfy the stated simplicial pseudomanifold conditions. In an $n$-cycle, cancellation at a common $(n-1)$-face determines the coefficient of either incident $n$-simplex from the other, up to the orientation sign. Connectivity through such faces therefore determines every top-simplex coefficient from one integer. If the signs are globally compatible, their oriented sum is a cycle and generates $H_n(K)\cong \mathbb{Z}$; if they are inconsistent, that integer must be zero and $H_n(K)=0$.

Let $x$ be the resulting fundamental class, represented by $z$. Since the intersection has dimension below $n$, split uniquely
$z=z_1+z_2$ into top chains in $L_1$ and $L_2$. Thenand the Mayer--Vietoris boundary isIt is nonzero exactly when both $L_1$ and $L_2$ contain top-dimensional simplices; if one side contains none, the fundamental cycle already lies in the other side, while if both do, connectedness forces a nonempty interface and its oriented boundary represents a nonzero class.

Finally take $K\cong S^3$, $L_1\cong S^1\times D^2$, and
$L_1\cap L_2\cong T^2$. The preceding boundaryis a nonzero map between copies of $\mathbb{Z}$ and sends the fundamental class to the oriented boundary torus, hence is an isomorphism. Exactness then givesThe next part of the sequence isSince $H_1(L_1)\cong \mathbb{Z}$, it follows that
$H_1(L_2)\cong \mathbb{Z}$. The degree-zero sequence makes $L_2$ connected. Therefore

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.