The Stacks project

Lemma 13.18.8. Let $\mathcal{A}$ be an abelian category. Let $I^\bullet $ be bounded below complex consisting of injective objects. Let $L^\bullet \in K(\mathcal{A})$. Then

\[ \mathop{\mathrm{Mor}}\nolimits _{K(\mathcal{A})}(L^\bullet , I^\bullet ) = \mathop{\mathrm{Mor}}\nolimits _{D(\mathcal{A})}(L^\bullet , I^\bullet ). \]

Proof. Let $a$ be an element of the right hand side. We may represent $a = \gamma \alpha ^{-1}$ where $\alpha : K^\bullet \to L^\bullet $ is a quasi-isomorphism and $\gamma : K^\bullet \to I^\bullet $ is a map of complexes. By Lemma 13.18.6 we can find a morphism $\beta : L^\bullet \to I^\bullet $ such that $\beta \circ \alpha $ is homotopic to $\gamma $. This proves that the map is surjective. Let $b$ be an element of the left hand side which maps to zero in the right hand side. Then $b$ is the homotopy class of a morphism $\beta : L^\bullet \to I^\bullet $ such that there exists a quasi-isomorphism $\alpha : K^\bullet \to L^\bullet $ with $\beta \circ \alpha $ homotopic to zero. Then Lemma 13.18.7 shows that $\beta $ is homotopic to zero also, i.e., $b = 0$. $\square$


Comments (2)

Comment #8417 by on

To justify "there exists a quasi-isomorphism with homotopic to zero," one could link the result I propose between parentheses in https://stacks.math.columbia.edu/tag/05RW#comment-8372

Comment #8828 by on

@#8785 Okay, I just realized: if has enough injectives, then the -functor is universal because it is erasable, i.e., it satisfies the hypothesis of 12.12.4: For an object , take an injection into an injective object. Then vanishes for all by 13.18.8. Dually, if has enough projectives, one can apply the analogous argument to to deduce erasibility (and thus universality).

Is the statement of this result to be found anywhere in the SP?

There are also:

  • 2 comment(s) on Section 13.18: Injective resolutions

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05TG. Beware of the difference between the letter 'O' and the digit '0'.