Kernels, images, quotients, and matrix coefficients of comodules

For the regular object take ${\rho }_C=\mathrm{\Delta }$. The coalgebra coassociativity identity is exactly $({\rho }_C\otimes {\mathrm{id}}_C){\rho }_C=({\mathrm{id}}_C\otimes \mathrm{\Delta }){\rho }_C$, and its two counit identities give $({\mathrm{id}}_C\otimes \epsilon ){\rho }_C={\mathrm{id}}_C$ (and the other counit identity), so $(C,\mathrm{\Delta })$ is a right comodule by the coalgebra definition [F2].

Let $M$ have ordered basis $v_1,\dots ,v_n$, and write ${\lambda }_i(x)$ for the unique coordinates of $x$ from [F8]. For $a,b\in k$ and $x,y\in M$, the coordinate list of $ax+by$ is $(a{\lambda }_i(x)+b{\lambda }_i(y){)}_i$ by the vector-space laws; uniqueness in [F8] therefore gives ${\lambda }_i(ax+by)=a{\lambda }_i(x)+b{\lambda }_i(y)$, so each coordinate function ${\lambda }_i:M\to k$ is linear. For each $j$, write ${\rho }_M(v_j)={\sum }_s{x}_s\otimes d_s$ as a finite tensor by [F3], expand each $x_s={\sum }_i{\lambda }_i(x_s)v_i$, and collect $c_{ij}:={\sum }_s{\lambda }_i(x_s)d_s$. Applying ${\lambda }_r\otimes {\mathrm{id}}_C$ recovers $c_{rj}$, so the expansion ${\rho }_M(v_j)={\sum }_i{v}_i\otimes c_{ij}$ is unique.

Let $K=\ker f$, $I=\mathrm{im}f$, and let $q:M\to M\mathrm{/}K$ be the quotient. The map $\bar{f}:M\mathrm{/}K\to N$, $\bar{f}(m+K)=f(m)$, is well-defined and injective: equal cosets have difference in $K$, and $\bar{f}(m+K)=0$ implies $m\in K$. It is linear because for $a,b\in k$ and cosets represented by $m,n\in M$, $\bar{f}(a(m+K)+b(n+K))=f(am+bn)=af(m)+bf(n)$. Thus $f=\bar{f}q$, so functoriality [F5] gives $f\otimes {\mathrm{id}}_C=(\bar{f}\otimes {\mathrm{id}}_C)(q\otimes {\mathrm{id}}_C)$. By [A1] and [F4], $\bar{f}\otimes {\mathrm{id}}_C$ is injective, while the kernel formula in [F4] with $U=K$ and $Z=0$ gives $\ker (q\otimes {\mathrm{id}}_C)=\mathrm{im}(K\otimes C)$. Therefore $\ker (f\otimes {\mathrm{id}}_C)=\mathrm{im}(K\otimes C\to M\otimes C)$. The same tensor-injection result makes the inclusion $K\otimes C\to M\otimes C$ injective.

Apply coassociativity to $v_j$ and use the expansion in 1.2 to get ${\sum }_{i,\mathrm{\ell }}{v}_i\otimes c_{i\mathrm{\ell }}\otimes c_{\mathrm{\ell }j}={\sum }_i{v}_i\otimes \mathrm{\Delta }(c_{ij})$. Applying ${\lambda }_r\otimes {\mathrm{id}}_C\otimes {\mathrm{id}}_C$ gives $\mathrm{\Delta }(c_{rj})={\sum }_{\mathrm{\ell }}{c}_{r\mathrm{\ell }}\otimes c_{\mathrm{\ell }j}$. Using the counit equation $({\mathrm{id}}_M\otimes \epsilon ){\rho }_M(v_j)=v_j$ under the tensor-unit identification and then applying ${\lambda }_r$ gives $\epsilon (c_{rj})={\delta }_{rj}$.

For $x\in K$, the comodule-map equation gives $(f\otimes \mathrm{id}){\rho }_M(x)={\rho }_N(f(x))=0$. By step 1.3, ${\rho }_M(x)$ has a unique preimage in $K\otimes C$ under the injective map $i\otimes \mathrm{id}$, where $i:K\hookrightarrow M$; define ${\rho }_K(x)$ to be that preimage. This is linear because it is the inverse of a linear isomorphism onto its image. The identity $(i\otimes \mathrm{id}){\rho }_K={\rho }_{M}i$ and the injectivity of $i\otimes \mathrm{id}\otimes \mathrm{id}$ show that coassociativity of ${\rho }_K$ follows from coassociativity of ${\rho }_M$; injectivity of $i$ and the same identity show its counit law. Hence $K$ is a subcomodule and $i$ is a comodule map.

Conversely, suppose the displayed matrix identities hold. Define $\rho$ on the basis by $\rho (v_j)={\sum }_i{v}_i\otimes c_{ij}$ and extend linearly using the unique coordinates from [F8]; the coordinate functionals are linear by step 1.2. For each basis vector, $(\rho \otimes \mathrm{id})\rho (v_j)={\sum }_{i,\mathrm{\ell }}{v}_i\otimes c_{i\mathrm{\ell }}\otimes c_{\mathrm{\ell }j}={\sum }_i{v}_i\otimes \mathrm{\Delta }(c_{ij})=(\mathrm{id}\otimes \mathrm{\Delta })\rho (v_j)$ by the coproduct identity; also $(\mathrm{id}\otimes \epsilon )\rho (v_j)={\sum }_i{v}_i{\delta }_{ij}=v_j$. The maps in each identity are linear and the basis spans $M$, so both identities hold on every vector; hence this is a right comodule.

Factor $f$ as $M\stackrel{q_I}{\to }I\stackrel{j}{\to }N$, where $I=\mathrm{im}f$ is a subspace by [F6], $q_I(m):=f(m)$ with codomain restricted to $I$ is the linear corestriction of $f$, and $j$ is inclusion. The map $q_I$ is onto by the definition of $I$. Define ${\rho }_I(q_I(m)):=(q_I\otimes \mathrm{id}){\rho }_M(m)$. If $q_I(m)=q_I(m^{\mathrm{\prime }})$, then $m-m^{\mathrm{\prime }}\in K$; by 2.2, ${\rho }_M(m-m^{\mathrm{\prime }})$ lies in the image of $K\otimes C$, which $q_I\otimes \mathrm{id}$ kills, so the formula is well-defined. It is linear and satisfies ${\rho }_I{q}_I=(q_I\otimes \mathrm{id}){\rho }_M$. Precomposing the coassociativity identity for ${\rho }_I$ with the surjection $q_I$ gives $(q_I\otimes \mathrm{id}\otimes \mathrm{id})({\rho }_M\otimes \mathrm{id}){\rho }_M=(q_I\otimes \mathrm{id}\otimes \mathrm{id})(\mathrm{id}\otimes \mathrm{\Delta }){\rho }_M$, so it holds by coassociativity on $M$ and surjectivity of $q_I$. Likewise, precomposing its counit identity with $q_I$ gives $q_I(\mathrm{id}\otimes \epsilon ){\rho }_M=q_I$, which holds by counitality on $M$; surjectivity gives the identity on $I$. Thus ${\rho }_I$ is a coaction. Since $f=j{q}_I$ is a comodule map, ${\rho }_{N}j{q}_I=(f\otimes \mathrm{id}){\rho }_M=(j\otimes \mathrm{id})(q_I\otimes \mathrm{id}){\rho }_M=(j\otimes \mathrm{id}){\rho }_I{q}_I$; surjectivity of $q_I$ proves ${\rho }_{N}j=(j\otimes \mathrm{id}){\rho }_I$, so $j$ is a comodule map.

Let $i:J_0\to M$ be an injective comodule map and set $J=i(J_0)$. By step 3.2, $J$ has the image coaction making its inclusion a comodule map. Let $\pi :M\to M\mathrm{/}J$ be the quotient and set $\bar{\rho }(m+J):=(\pi \otimes \mathrm{id}){\rho }_M(m)$. If $m$ is replaced by $m+j$, the difference is $(\pi \otimes \mathrm{id}){\rho }_M(j)=0$ because ${\rho }_M(j)$ is the image of a tensor in $J\otimes C$; hence $\bar{\rho }$ is well-defined. It is linear and $\bar{\rho }\pi =(\pi \otimes \mathrm{id}){\rho }_M$. Precomposing its coassociativity identity with the surjection $\pi$ gives $(\pi \otimes \mathrm{id}\otimes \mathrm{id})({\rho }_M\otimes \mathrm{id}){\rho }_M=(\pi \otimes \mathrm{id}\otimes \mathrm{id})(\mathrm{id}\otimes \mathrm{\Delta }){\rho }_M$, so it holds by coassociativity on $M$ and surjectivity of $\pi$. Precomposing its counit identity with $\pi$ gives $\pi (\mathrm{id}\otimes \epsilon ){\rho }_M=\pi$, which follows by counitality on $M$; surjectivity gives the identity on $M\mathrm{/}J$. Thus the quotient is a comodule and $\pi$ is a comodule map.

The map $\tilde{f}:M\mathrm{/}K\to I$, $m+K\mapsto f(m)$, is the same induced map as $\bar{f}$ with codomain restricted to $I$. It is injective by the definition of $K$ and onto by the definition of $I$, hence is a vector-space isomorphism. The quotient and image coaction formulas in steps 3.2 and 4.1 give ${\rho }_I\tilde{f}=(\tilde{f}\otimes \mathrm{id})\bar{\rho }$. Since $\tilde{f}$ and its tensor map are isomorphisms, composing this identity with their inverses shows ${\tilde{f}}^{-1}$ is also a comodule map.

Step 1.1 establishes the regular comodule; steps 1.2, 2.1, and 3.1 prove both directions of the finite coefficient-matrix characterization. Steps 1.3, 2.2, 3.2, 4.1, and 5.1 establish the kernel, image, quotient, and first-isomorphism claims under AC.