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There is the statement on page $652$ in Chapter $10.5$ on the subject of Cartan Eilenberg Resolution: "($C$ is a chain complex) given an object $A'$, let $Q^n (A')$ be the complex with $A'$ concentrated in degree $n$; given a morphism $f:C_n \rightarrow A'$, define a chain map $F=\left( F_i \right) : C_n \rightarrow Q^n (A')$, where $F_n = f$ and all the other $F_i = 0$." But unfortunately, $F$ is not a chain map! Am I right on this?

the related picture is: $Theorem$ $10.42$ in Rotman's AIHA book

fufufufuf
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    It's fine to include an image, but it might be a good idea to also write out the relevant text/statement out in the body of your question. – Christian E. Ramirez Feb 18 '23 at 04:53
  • Welcome to math.SE! Please see https://math.meta.stackexchange.com/questions/34121/why-image-cannot-be-used-for-explaining-my-maths-problem/34123#34123. – joriki Feb 18 '23 at 05:55
  • I seem to agree. I do not see a reason why this $F$ should be a chain map. – Jochen Feb 18 '23 at 11:39
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    F will be a complex map exactly when the map f is a cocycle. – Mariano Suárez-Álvarez Feb 19 '23 at 03:30
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    Why would providing an actual image of the text about which the question is about be lack of context and detail? Yes, it is better if the image is embedded in the question, but the user cannot do that because of low reputation, and instead of voting to close without even explaining the reason literally anyone could have edited the image in. This is an user who has used the site for two days. And, mind you, the question is a good one. – Mariano Suárez-Álvarez Feb 19 '23 at 03:35
  • For a proof of what you want, see https://math.stackexchange.com/a/799140/274 (of which this is not a dupe, as this question is about a bug in a specific book) @fufufufuf you should look for a concrete, small example in which the map is not a map of complexes, and write an answer to your own question (and spell out the book's title!) – Mariano Suárez-Álvarez Feb 19 '23 at 03:50
  • @MarianoSuárez-Álvarez Thanks – fufufufuf Feb 19 '23 at 03:58
  • @MarianoSuárez-Álvarez Now I see Rotman is half way there.We have $C$ is split,thus $C$=$Z$⊕$B$$[$-1$]$,it suffices to prove every term of $Z$ and $B$$[$-1$]$ are projective. and the direct summand of a projective is projective, thus $Z$ and $B$$[$-1$]$ are projective. and $B$$[$-1$]$ is projective, so is $B$$[$+1$]$. For $Z$ and $B$$[$+1$]$,the chain map $F$ is a proper chain map, hence every term of $Z$ and $B$$[$+1$]$ are projective. – fufufufuf Feb 19 '23 at 08:55

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Assume for simplicity that we're working over Abelian groups, and consider a chain complex $C$ of the form $$ \cdots \stackrel{d_{n+1}}\longrightarrow C_n\stackrel{d_n}\longrightarrow C_{n-1}\stackrel{d_{n-1}}\longrightarrow\cdots $$ and an Abelian group $A$, and let $A[n]$ denote the chain complex concentrated in degree $n$ with $A[n]_n = A$. Then a map $F : C \longrightarrow A[n]$ of chain complexes must satisfy that $dF(c) = F(dc)$ for all $c\in C$. Since $F$ is the same datum as a map of Abelian groups $f : C_n \to A_n$ and since $A[n]$ has trivial differential, you see that the condition is that for all $c\in C_{n+1}$, one has that $f(dc) = 0 $.

In other words, $f$ must vanish on the $n$ boundaries of $C_n$, so $f$ induces a map $\tilde f: \operatorname{coker} d_{n+1} \to A$. Thus, the functor $A\mapsto A[n]$ is actually right adjoint to the functor $(C,d) \mapsto \operatorname{coker} d_{n+1}$. You can check that $A\mapsto A[n]$ is also left adjoint to the functor $(C,d) \mapsto \ker d_n$. Thus, things are a little more complicated than what is suggested by the book.

Rotman's book on homological algebra has many typos and errors. I suggest you consult a list of errata, such as this, though I do not find this mistake there quickly scanning it. Depending on the edition, there are entire passages or proofs that have to be replaced.

Pedro
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