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Let ‎$‎‎\mathcal{C}‎$ ‎be a ‎closed ‎symmetric ‎monoidal‎ ‎Grothendieck ‎category. ‎Then ‎there ‎are ‎two ‎general ‎notions ‎of ‎purity ‎in ‎‎$‎‎\mathcal{C}‎$‎, ‎the ‎‎$‎‎\lambda‎‎$‎-purity and the ‎$‎\otimes‎$‎-purity. ‎‎ ‎ Let ‎$‎f:A\to B‎$ ‎be a‎ ‎morphism ‎in ‎‎$‎‎\mathcal{C}‎$‎. ‎‎$‎f‎$ ‎is ‎called ‎‎$‎‎\lambda‎‎$‎-pure if for any commutative diagram‎$\require{AMScd}$ ‎\begin{CD} A' @>f'>> B' \\ @VuVV @VVvV \\ ‎A @>>f> B \end{CD} there exists a map ‎$‎g:B'\to A‎$ ‎such ‎that ‎‎$‎u=g\circ f'‎$‎. One can see that every ‎$‎\lambda‎$‎-pure morphism is a monomorphism.‎ ‎ A monomorphism ‎$‎f:X\to Y‎$ ‎is ‎said ‎to ‎be ‎‎$‎\otimes‎$‎-pure if for any object ‎$‎Z‎$ ‎in ‎‎$‎‎\mathcal{C}‎$‎, ‎‎$f\otimes ‎Z‎$ ‎is ‎monic.‎ One ‎can ‎see ‎that ‎every ‎‎$‎\lambda‎$‎-pure monomorphism is ‎$‎\otimes‎$‎-pure. ‎ ‎ Now let us consider the category of chain complexes of ‎$‎‎\mathcal{C}‎$ ‎and ‎denote ‎it ‎by ‎‎$‎‎‎\mathbb{C}(‎\mathcal{C}‎)‎$‎.

How can I define the notion of purity in $\mathbb{C}(\mathcal{C})$?

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  • $\begingroup$ I cannot read your diagram. Could you please check that the LaTeX is correct? $\endgroup$ – Jason Starr Jul 12 '16 at 12:32
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    $\begingroup$ The flavor of LaTeX supported here doesn't include xymatrix, unfortunately, nor does it care about square brackets. $\endgroup$ – Qiaochu Yuan Jul 12 '16 at 18:51
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    $\begingroup$ I might be missing something, but it seems to me that $\mathbb{C}(\mathcal{C})$ is again a Grothendieck category, with a closed symmetric monoidal structure $\otimes$ defined using the $\otimes$ on $\mathcal{C}$ just as it's defined for chain complexes of modules (this can also be viewed as an enriched Day convolution, I think). So both definitions of purity that you gave also make sense in $\mathbb{C}(\mathcal{C})$. Is your question really "how is purity in $\mathbb{C}(\mathcal{C})$ related to purity in $\mathcal{C}$"? $\endgroup$ – Tim Campion Jul 14 '16 at 17:33
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    $\begingroup$ Also, in the definition of $\lambda$-pure, I think you mean to require that $A'$ and $B'$ are $\lambda$-presentable, right? Otherwise the cardinal $\lambda$ doesn't enter in at all... $\endgroup$ – Tim Campion Jul 14 '16 at 17:37

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