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Thank you for your answer! I'm confused by your last formula, $S^n\otimes_A B \simeq S^n \odot (B\wedge_A B)$, I don't see how it follows from what you wrote.
@AaronMazel-Gee If $*$ is a zero object (is this what you meant by “empty”?), then a similar naïve definition of a $Q’$ as above also won’t work. Again, consider $X\to *\to Y$; on one hand it should go to $QX\to Q*\to QY$, and on the other hand it should go to $QX\to *\to QY$. This leads to exactly the same problem as before. What does the little triangle mean in your displayed arrow?
@HarryGindi (cont.) Indeed, consider a composition $X\stackrel{f}{\to} *\stackrel{g}{\to} Z$ where $X$ and $Z$ are not $*$. Then $Q’$ maps the composition to $QX\to Q*\to QZ$. On the other hand, the composition of the $Q’$’s of these arrows is $QX\to Q*\to *\to QZ$. So now the question is whether $Qg:Q*\to QZ$ is equal to $Q*\to *\to QZ$, which is a case of assertion (A) above, which doesn’t hold for $Q$ (if it did, there would be no need to go through all this).
@AaronMazel-Gee I don't see how this would work. For simplicity, suppose instead of modifying the whole functorial factorization we just modify the functor $Q$, so we ask ourselves whether we can construct $Q'$, an endofunctor of $\mathcal C$ such that $Q’X$ is $QX$ if $X\not= *$ and is $*$ if $X=*$. OK, then on an arrow $X\to Y$ it should be $QX\to QY$ if $X,Y\not= *$, it should be $*\to QY$ if $X=*$, and it should be $QX\to Q*\to *$ if $X\not=*$ and $Y=*$. But this is not always a functor.
