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I am working through computing the homotopy of Thom spectra from Kochman's book. Let $A$ be a coalgebra over a field $k$, and let $M$ be a right $A$-comodule. Kochman constructs a coresolution $F$ of $M$ as follows: Define $F_0=M \otimes A$ and $\eta_0=\psi_M$ (the coaction of $A$ on $M$). Inductively, if $F_n$ and $\eta_n: K_n \rightarrow F_n$ have been defined, we define $K_{n+1}$ to be the cokernel of $\eta_n$. Then we define $F_{n+1}=K_{n+1} \otimes A$ and $\eta_{n+1}=\psi_{K_{n+1}}$ (the coaction of $A$ on $K_{n+1}$).

We then get short exact sequences $$0 \rightarrow K_n \rightarrow F_n \rightarrow K_{n+1} \rightarrow 0$$ Which we may splice together into a coresolution of $M$ by the $F_n$. Furthermore, this coresolution has the form $F=F' \otimes A$, which is important later when Kochman proves a change of rings proposition.

Question: In this construction, why does $F_n=K_n \otimes A$ have to be free?

$K_n$ is the cokernel of $\eta_n: K_n \rightarrow K_n \otimes A$, and it is not clear to me why this cokernel should be a direct sum of copies of $A$. I am not very comfortable working with comodules and coresolutions, so any help here will be greatly appreciated.

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Let $V$ be a vector space. Then $V\otimes A$ is a comodule under the coaction $V\otimes A\rightarrow (V\otimes A)\otimes A$ given by tensoring $V$ with the comultiplication $A\rightarrow A\otimes A$. If we can show that $V\otimes A$ is a direct sum of copies of the comodule $A$, then taking $V=K_n$ will show $F_n$ is free.

First note that since V is a vector space, $V\cong \bigoplus_{\alpha} k $ where $\alpha$ ranges over some indexing set. Without loss of generality, we may assume that $V=\bigoplus_{\alpha}k$. Then since the tensor product preserves colimits in $Vect$, we have $\bigoplus_{\alpha} A\cong\bigoplus_\alpha k\otimes A\cong(\bigoplus_{\alpha}k)\otimes A=V\otimes A $ as vector spaces. The left side is a comodule via the map $\bigoplus_{\alpha}A\rightarrow\bigoplus_{\alpha} A\otimes A\cong (\bigoplus_{\alpha}A)\otimes A$ given by comultiplying each summand.

All that remains is to show that the obvious map $\bigoplus_\alpha A\rightarrow V\otimes A$ is a comodule homomorphism, so that it will be a comodule isomorphism. In other words, we must check the commutativity of a square involving this map, it's tensor product with $A$, and the coactions on both comodules. This is easy to see by applying the naturality of the obvious transformation $\bigoplus_\alpha X\rightarrow (\bigoplus_\alpha k)\otimes X$ to the comultiplication $A\rightarrow A\otimes A$. The resulting commutative square will be the one we want since the coaction on each comodule is induced by comultiplication; the only difficulty is to check that the map $\bigoplus_\alpha A\otimes A\rightarrow (\bigoplus_\alpha k)\otimes A\otimes A$ coming from the natural transformation agrees with the one given by tensoring the map $\bigoplus_\alpha A\rightarrow V\otimes A$ with $A$. However, it is easy to check both maps agree on elements of the basis of $\bigoplus_\alpha A\otimes A$ induced by a basis of $A$.

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Thank you for the great answer! I didn't realize that since the coaction on $V \otimes A$ is just the comultiplication on the second component (which is the same as the coaction on $V$ tensored with the identity), we may replace $V$ with the isomorphic vector space $\oplus_\alpha k$. –  Vitaly Lorman Dec 29 '11 at 17:58
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