I am struggling to show that the straightening construction (2.2.1 in Higher Topos Theory) preserves colimits. More specifically let $M_X:=\mathfrak{C}X^\triangleright\sqcup_{\mathfrak{C}X}C^{op}$ for $X\to S$ a simplicial set over $S$ and $\phi\colon\mathfrak{C}S\to C^{op}$ a functor between simplicially enriched categories. If $\infty\in M_X$ denotes the image of the cone point of $\mathfrak{C}X^\triangleright$ and if $i\colon C^{op}\to M_X$ is the inclusion then the straightening of $X$ is defined as $St_\phi X=map_{M_X}(i(-),\infty)$. Showing that this functor preserves colimits provides us with its right adjoint, the unstraightening.

I can think of two approaches to this:

  1. Let $X\colon I\to(Set_{\Delta})_{/S}$ with $i\mapsto X_i$ be a small diagram. Then $colim_i(St_\phi X_i)\cong St_\phi(colim_i X_i)$ would follow if we showed that $colim_i M_{X_i}\simeq M_{colim_i X_i}$. The problem is that the cone functor $X\mapsto X^\triangleright$ only preserves connected colimits and so $M_{(-)}$ probably doesn't preserve small colimits (I think it is not to hard to construct a counterexample by letting $\phi$ be the identity, $S=\Delta^0$ and $X=\Delta^0\sqcup\Delta^0$ be a coproduct of two points).
  2. Another appraoch would be to directly construct the unstraightening. We know how this is supposed to look by using our knowledge about a left Kan extension and its right adjoint along the Yoneda embedding. However that the construction we get is really a right adjoint to the straightening defined above is not easy to see and doesn't seem to be the right approach to this problem.

Any hints or thoughts are welcome.

edit: My third approach was wrong.


1 Answer 1


Okay so the answer seems to be easier than I initially thought, because I kept thinking about the straightening of coproducts. A functor preserves all small colimits if and only if it preserves the initial object and wide pushouts. It is not hard to see that the straightening of the empty simplicial set is the functor $C\to Set_{\Delta}$ that is constant on the empty set. Now joins and in particular the functor $-\star\Delta^0$ preserve connected colimits (as an endofunctor on $Set_\Delta$). Since pushouts and $\mathfrak{C}$ commute with colimits we deduce that the assignment $X\mapsto M_X$ preserves wide pushouts (because they are always connected). Thus straightening also preserves wide pushouts. This proves the claim.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.