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I'm trying to solve the following problem:

Suppose that $\mathcal{A}$ is an abelian category which has enough projective objects (i.e. for every $A\in \mathcal{A}$ there is $P\in\mathcal{A}$ projective and an epimorphism $P\to A\to 0$). Let $\mathcal{Ch(A)}$ be the category of cochains in $\mathcal{A}$. Prove that $\mathcal{Ch(A)}$ has enough projective objects.

Remember the following lemma: $P^{\circ}$ is projective in $\mathcal{Ch(A)}$ $\iff$ $P^n$ is projective $\forall n$ and $P^{\cdot}$ is contractible (i.e. both exact and split).

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Hint: The building blocks for the projectives in $\text{Ch}({\mathscr A})$ are the complexes $\ldots\to 0\to P\to P\to 0\to\ldots$ with $P\in\text{Proj}({\mathscr A})$. Also, consider the case of a stalk complex $\ldots\to 0\to X\to 0\to\ldots$ first.

Some more detail as requested: The forgetful functor ${\mathsf V}:\text{Ch}({\mathscr A})\to {\mathscr A}^{\mathbb Z}$ has a left adjoint ${\mathsf C}: {\mathscr A}^{\mathbb Z}\to \text{Ch}({\mathscr A})$ sending $X\in{\mathscr A}$ to $\ldots\to 0\to X\to X\to 0\to\ldots$. As the left adjoint to the exact and faithful functor ${\mathsf V}$, ${\mathsf C}$ preserves projectives and epimorphisms, and moreover the counit $\eta: {\mathsf C}\circ{\mathsf V}\to\text{id}$ is an epimorphism (since ${\mathsf V}\eta$ is a split epimorphism). This reduces the existence of enough projectives from $\text{Ch}({\mathscr A})$ to ${\mathscr A}^{\mathbb Z}$.

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  • $\begingroup$ Hi. Thank you for the hint. Can you help me also for the following? math.stackexchange.com/questions/1926554/… $\endgroup$
    – DDT
    Sep 17, 2016 at 12:55
  • $\begingroup$ @DDT: Yep, I tried! $\endgroup$
    – Hanno
    Sep 17, 2016 at 13:05
  • $\begingroup$ @Hanno Could I ask for a further hint? The case of a stalk complex is easy, it's extending to arbitrary complexes that is giving me trouble. $\endgroup$
    – D. Brogan
    Apr 23, 2020 at 22:33
  • $\begingroup$ @D.Brogan Sure, I've amended the answer. Let me know if have any questions. $\endgroup$
    – Hanno
    Apr 24, 2020 at 6:22
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    $\begingroup$ @Hanno You've defined $C:\mathscr A^{\mathbb Z}\to\mathrm{Ch}(\mathscr A)$ for $X\in\mathscr A$... How do you extend to $\mathscr A^{\mathbb Z}$? I want to take the sum of complexes but I don't think this works... $\endgroup$
    – D. Brogan
    May 1, 2020 at 19:59
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Obviously each cyclic complex has a projective resolution. For $C_* \in Ch(A)$, define $K_n=\ker(C_n\to C_{n-1})$, $I_n=\mathrm{Im}(C_{n+1}\to C_n)$. Let $K_*$, $I_*$ be the corresponding cyclic complexes. Apply Horseshoe lemma to the short exact sequence \begin{equation} 0\to K_* \to C_* \to I_*[-1]\to 0.\end{equation}

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If i have $A\in\mathcal{Ch(A)}$, i can find for each $n\in\mathbb{Z}$ an epimorphism $\gamma^n:P^n\to A^n$. Let $d_A^n:A^n\to A^{n+1}$ so we have also morphisms $d_A^n\circ\gamma^n:P^n\to A^{n+1}$. By projective property we can find $\alpha^n:P^n\to P^{n+1} $ s.t. $d_A^n\circ \gamma^n=\gamma^{n+1}\circ\alpha^n$

But the discovered $\alpha^n$ doesn't give a cochain. So using the ideas of the building blocks we consider the following chain $P^n\oplus P^{n+1}\to P^{n+1}\oplus P^{n+2}$ with maps $\left[\begin{matrix}\alpha^n &id_{P^{n+1}}\\ 0&0\end{matrix}\right]$ It's split and exact by building. We map it to $A^{\cdot}$ in such way: $[\gamma^n\;\;0]:P^n\oplus P^{n+1}\to A^n$. It is a map of cochain

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  • $\begingroup$ Your answer is wrong. Your differentials are not differentials. $\endgroup$
    – XT Chen
    Oct 25, 2019 at 13:45

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