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Jul 11, 2010 at 11:32 comment added Tom Goodwillie Oh, good. I was almost ready to sign myself "enriched, tensored, and bewildered".
Jul 11, 2010 at 5:07 vote accept Harry Gindi
Jul 11, 2010 at 5:07 comment added Harry Gindi Alright, I've e-mailed Lurie, and he said that the statement of the proposition is incorrect (and hence your counterexample works)!
Jul 11, 2010 at 4:21 comment added Harry Gindi It's tensored if for every C-object X and V-object K, $Map(A,-)^K\cong Map(B,-)$ for some object $B$ in $C$. (This may be identical to what you said by some Yoneda thing, but to be honest, I did not check).
Jul 11, 2010 at 3:29 comment added Tom Goodwillie Don't take my word for it. Do I have the definitions right? I think that if $C$ is enriched over $V$ then it's said to be tensored if for every $C$-object $A$ the functor $Map(A,-)$ from $C$ to $V$ has a left adjoint; and cotensored if for every $C$-object $B$ the functor $Map(-,B)$ from $C$ to $V^{op}$ has a right adjoint.
Jul 11, 2010 at 2:23 comment added Tom Goodwillie I didn't interpret "living in $L$" as embedding in $L$. But let's make it $\Delta^1\leftarrow\partial\Delta^1\to\Delta^1$ instead.
Jul 11, 2010 at 2:17 comment added Harry Gindi well, the example you gave doesn't fit exactly, since both $\partial\Delta^1$ should embed in both $L$ and $\Delta^1$. That wouldn't stop you from coming up with another counterexample, of course.
Jul 11, 2010 at 1:48 history answered Tom Goodwillie CC BY-SA 2.5