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Well my question is as clear as its title suggests. So here I would like to clarify on the fact that an object $A^\cdot$ in $\mathrm{Ch}^{\geq 0}(\mathcal{A})$ is injective if and only if $0\longrightarrow\mathrm{H}^0(A^\cdot)\longrightarrow A^0\longrightarrow A^1\longrightarrow\cdot\cdot\cdot$ is exact and $\mathrm{H}^0(A^\cdot)$ and all $A^n$'s are injectives in $\mathcal{A}$.

Many thanks in advance!

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    $\begingroup$ The title question is related to this one, but the body question seems to be something different. $\endgroup$
    – Zhen Lin
    Commented Jul 30, 2014 at 13:15
  • $\begingroup$ I think you need more, namely that $\mathrm{Ker}\, d^i$ is a direct factor of $A^i$ for each $i\geq 0$. See A. Grothendieck, Sur quelques points d'algèbre homologique, p. 146. $\endgroup$
    – abx
    Commented Jul 30, 2014 at 13:30
  • $\begingroup$ @abx I think that follows from the other conditions. By induction on $i$, $\operatorname{ker}(d^i)$ is injective and so $0\to\operatorname{ker}(d^i)\to A^{i+1}\to\operatorname{ker}(d^{i+1})\to0$ splits. $\endgroup$
    – Jeremy Rickard
    Commented Jul 30, 2014 at 14:07
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    $\begingroup$ By the Dold-Kan correspondence, the category of chain complexes is a presheaf category. $\endgroup$
    – John Wiltshire-Gordon
    Commented Jul 30, 2014 at 14:53
  • $\begingroup$ @Jeremy Rickard: Right! Strange that Grothendieck doesn't put it that way. $\endgroup$
    – abx
    Commented Jul 30, 2014 at 14:54

1 Answer 1

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If $X^{\bullet}$ is an object of $\mathrm{Ch}^{\geq0}(\mathcal{A})$ and for each $i\geq0$ $X^i\to I^i$ is an embedding into an injective, then $X^{\bullet}$ embeds in $$I^{\bullet}:=\dots\to 0\to I^0\oplus I^1\to I^1\oplus I^2\to I^2\oplus I^3\to\dots$$ (with the obvious maps as differentials). $I^{\bullet}$ is the product (and coproduct) of cochain complexes $J_0^{\bullet}:=\dots0\to I^0\to0\to\dots$ and $J_i^{\bullet}:=\dots0\to I^i\to I^i\to0\to\dots$ for $i>0$. Since products of injective objects are injective, it suffices to prove that $J_i^{\bullet}$ is injective for all $i\geq0$, which is straightforward since for any object $Y^{\bullet}$ of $\mathrm{Ch}^{\geq0}(\mathcal{A})$, $\operatorname{Hom}(Y^{\bullet},J_i^{\bullet})=\operatorname{Hom}_{\mathcal{A}}(Y^i,I^i)$.

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    $\begingroup$ In Jeremy's answer, the embedding is $(\eta^n,f^n\circ\eta^n)^T:X^n\to I^n\oplus I^{n+1}$ where $\eta^n$ is the embedding $X^n\to I^n$ and $f^n:I^n\to I^{n+1}$ is an extension using injectivity of $I^{n+1}$. Might we need not (co)product of infinitely many complexes which imposes (AB3)* or (AB3) on $\mathcal{A}$, but only the criterion for injectivity in $\mathrm{Ch}^{\geq 0}(\mathcal{A})$ given in the OP? E.g., for $\mathrm{Ch}(\mathcal{A})$, $A^\cdot$ is injective iff $A^\cdot$ is exact and $A^n$'s and $\ker d^n$'s are injective. So the construction would apply to $\mathrm{Ch}(\mathcal{A})$. $\endgroup$
    – W.Z.
    Commented Jul 31, 2014 at 2:23
  • $\begingroup$ In my answer, the infinite products were of families of complexes such that in each degree only finitely many of the complexes are non-zero, which doesn't need AB3*. $\endgroup$
    – Jeremy Rickard
    Commented Jul 31, 2014 at 8:21
  • $\begingroup$ It's not true in general that an object of $\mathrm{Ch}(\mathcal{A})$ is injective if it's exact and each term is injective (which seems to have been introduced into your post in an edit by somebody else): you need $\operatorname{ker}(d^i)$ to be a direct factor of the degree $A^i$ (which is not automatic for complexes unbounded to the left). $\endgroup$
    – Jeremy Rickard
    Commented Jul 31, 2014 at 8:27
  • $\begingroup$ Yeah, I think you're right, we need each $\mathrm{ker}d^i$ to be injective or for the case of $\mathrm{Ch}^{\geq 0}(\mathcal{A})$ it is required to have $\mathrm{H}^0(A^\cdot)$ to be injective. Actually I introduced this mistaken condition but somehow it was credited as Community. Apology for this mathematical mistake! $\endgroup$
    – W.Z.
    Commented Jul 31, 2014 at 10:53
  • $\begingroup$ To be clear, in $\mathrm{Ch}^{\geq 0}(\mathcal{A})$, if $\mathrm{H}^0(A^\cdot)$ is injective and $0\longrightarrow\mathrm{H}^0(A^\cdot)\longrightarrow A^0\longrightarrow A^1\longrightarrow\cdot\cdot\cdot$ is exact then $0\longrightarrow\mathrm{H}^0(A^\cdot)\longrightarrow A^0\longrightarrow\mathrm{ker}d^1\longrightarrow 0$ is split exact. Since we require $A^n$'s to be injective, so is $\mathrm{ker}d^1$. By induction, we have all injective $\mathrm{ker}d^n$'s. $\endgroup$
    – W.Z.
    Commented Jul 31, 2014 at 10:59

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