Theorem: Exact descent criterion

There exists a unique linear map

satisfying

if and only if

If ${\mathcal{K}}_{\mathrm{int}}$ carries the quotient topology induced by $\mathcal{P}/\ker \mathsf{Rec}$, a continuous $\tilde{\mathsf{M}}$ satisfying (eq:descent) induces a continuous $\mathsf{M}$.

Proof. If the factorization exists, every null presentation is mapped to zero.

Conversely, if $\mathsf{Rec}A=\mathsf{Rec}{A}^{\prime }$, then $A-A^{\prime }\in \ker \mathsf{Rec}$. Condition (eq:descent) gives $\tilde{\mathsf{M}}(A)=\tilde{\mathsf{M}}(A^{\prime })$. Thus

is well-defined and linear. Surjectivity onto ${\mathcal{K}}_{\mathrm{int}}$ gives uniqueness. Continuity is the defining universal property of the quotient topology.