2
$\begingroup$

Let $D(A)$ be the derived $(\infty,1)$-category of some abelian category $A$. For which $A$ is $D(A)$ locally cartesian closed?

Replace $D$ with $D^b$ or similar if appropriate.

I essentially want to show that for any $f:x\to y$ in $D(A)$, the pullback functor $f^*: D(A)/y\to D(A)/x$ has a right adjoint $f_*$. However I am having trouble imagining what $f_*$ would look like in a derived category $D(A)$.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

This will almost never happen. Since $D(A)$ has a terminal object 0, if it's locally cartesian closed, then it's also cartesian clsoed. To be cartesian closed means that $x \oplus (-) : D(A) \to D(A)$ is a left adjoint, and in particular preserves colimits. In particular, it preserves the initial object: $x \oplus 0 = 0$. So this only happens when $D(A)$ is the terminal category.

All we used here is that $D(A)$ is (semi)additive, and this is true in both ordinary category theory and in $\infty$-category theory.

Improve this answer
$\endgroup$
0

You must log in to answer this question.