Let $\mathscr{C}$ be a pointed $\infty$-category. Then, $\Sigma : \mathscr{C} \rightleftarrows \mathscr{C} : \Omega$ can be defined through $\Sigma = \operatorname{colim}( * \leftarrow X \to *)$ and $\Omega = \lim ( * \to X \leftarrow *)$.
There seems to be some part in the theory of (co-)limits and adjunctions for $\infty$-categories that I'm not understanding well enough. I wanted to verify $\Sigma \dashv \Omega$ which caused me a lot of confusions. Initially, I thought that this would be a quick computation via \begin{align*} \operatorname{map}(\Sigma x, y) &\simeq \operatorname{map}(*, y) \times_{\operatorname{map}(x,y)} \operatorname{map}(*, y) \\ &\simeq \Omega \operatorname{map}(x,y) \\ &\simeq \operatorname{map}(x,*) \times_{\operatorname{map}(x,y)} \operatorname{map}(x,*) \\ &\simeq \operatorname{map}(x, \Omega y). \end{align*}
This seems fine but to show that we get an adjunction one should really prove that $$ \operatorname{map}(\Sigma x, y) \xrightarrow{\Omega} \operatorname{map}(\Omega \Sigma x, \Omega y) \xrightarrow{\eta^*} \operatorname{map}(x, \Omega y)$$ is an equivalence where $\eta$ is induced by the universal property of $\Omega$.
I thought this was fine: We can just write out suitable comparison maps to the top equivalence chain and call it a day. However, I cannot manage to do so. Could someone spell out the details of checking that really this composite $\eta^* \circ \Omega$ yields an equivalence of mapping anima?
Best Answer
The functor $\Omega$ is the composite of a bunch of functors, which I will now describe. Write $\mathsf{A}$ for the diagram $2\to 3\leftarrow 1$. The inclusion $i_3\colon *\to\mathsf{A}, *\mapsto 3$ induces a left Kan extension $\mathrm{Lan}_{i_3}\colon\mathscr{C}\to\mathrm{Fun}(\mathsf{A},\mathscr{C})$ (informally sending $X\in\mathscr{C}$ to $*\to X\leftarrow*$). Write $\mathsf{Sq}$ for the category $[1]\times[1]$, but instead of $(00)$, $(01)$, $(10)$, and $(11)$, I call the objects $0$, $1$, $2$ and $3$, respectively. Write $i\colon\mathsf{A}\to\mathsf{Sq}$ for the obvious inclusion. This induces a right Kan extension $\mathrm{Ran}_i\colon\mathrm{Fun}(\mathsf{A},\mathscr{C})\to\mathrm{Fun}(\mathsf{Sq},\mathscr{C})$. Finally, the inclusion $i_0\colon *\to\mathsf{Sq}, *\mapsto 0$ induces a restriction functor $i_0^*\colon\mathrm{Fun}(\mathsf{Sq},\mathscr{C})\to\mathscr{C}$. The functor $\Omega$ is the composite $$ \mathscr{C}\xrightarrow{\mathrm{Lan}_{i_3}}\mathrm{Fun}(\mathsf{A},\mathscr{C})\xrightarrow{\mathrm{Ran}_i}\mathrm{Fun}(\mathsf{Sq},\mathscr{C})\xrightarrow{i_0^*}\mathscr{C}. $$ The functor $i_0^*$ is equivalent to the limit functor $\mathrm{lim}\colon\mathrm{Fun}(\mathsf{Sq},\mathscr{C})\to\mathscr{C}$ (since the limit of a diagram with an initial object equals the value at this initial object), so $i_0^*$ has the constant diagram functor $\Delta\colon\mathscr{C}\to\mathrm{Fun}(\mathsf{Sq},\mathscr{C})$ as left adjoint. The functor $\mathrm{Ran}_i$ is right adjoint to $i^*\colon\mathrm{Fun}(\mathsf{Sq},\mathscr{C})\to\mathrm{Fun}(\mathsf{A},\mathscr{C})$ by definition. Finally, we claim that $\mathrm{Lan}_{i_3}$ is also a right adjoint functor. Let us find a left adjoint.
Write $\mathsf{B}$ for the category $(-2)\leftarrow 2\to 3\leftarrow 1\to (-1)$. There is an inclusion $j\colon\mathsf{A}\to\mathsf{B}, n\mapsto n$. Let $F\colon\mathrm{Fun}(\mathsf{A},\mathscr{C})\to\mathscr{C}$ be the composite $$ \mathrm{Fun}(\mathsf{A},\mathscr{C})\xrightarrow{\mathrm{Ran}_j}\mathrm{Fun}(\mathsf{B},\mathscr{C})\xrightarrow{\mathrm{colim}}\mathscr{C}. $$ Informally, $F$ sends a diagram $A\to C\leftarrow B$ to $\mathrm{colim}(*\leftarrow A\to C\leftarrow B\to *)$. (This functor is informally the ''main reason'' that the left adjoint to $\Omega$ will be the suspension functor).
We claim that $F$ is left adjoint to $\mathrm{Lan}_{i_3}$. Given a functor $X\colon\mathsf{A}\to\mathscr{C}$, and an object $Y\in\mathscr{C}$, the mapping anima $\mathscr{C}^\mathsf{A}(X,\mathrm{Lan}_{i_3}Y)$ (an ''anima'' is a synonym for ''homotopy type'' that the cool kids sometimes use nowadays --I use it because Qi did, not because I am one of those cool kids myself) can be pictured as the anima of diagrams
where the dashed maps may be freely picked, and the solid arrows are fixed. Let $p_1\colon X_1\to *$, $p_2\colon X_2\to *$, $p_{31}\colon X_3\to*$ and $p_{32}\colon X_3\to *$ be the maps given to us in the diagram $\mathrm{Ran}_j X$, which come with homotopies $p_1\simeq(X_1\to X_3\xrightarrow{P_{31}}*) and $p_2\simeq(X_2\to X_3\xrightarrow{p_{32}}*). The anima of maps $X\to *$ is contractible, as is the anima of maps $*\to Y$. Via a pullback of the anima of diagrams as above along the inclusion of those diagrams where the maps $X_1\to *$, $X_2\to *$, $X_3\rightrightarrows *$ and the homotopies between them are given by $p_1$, $p_2$, $p_{31}$, $p_{32}$ and the homotopies between them, we get a naturally(!) equivalent anima of diagrams
Likewise, we have a natural equivalence from the above anima towards the anima of diagrams
where we no longer require the maps $*\rightrightarrows Y$ to be specified. This anima is naturally equivalent to the anima $\mathscr{C}^\mathsf{B}(\mathrm{Ran}_j X, \Delta Y)$, with $\Delta\colon\mathscr{C}\to\mathrm{Fun}(\mathsf{B},\mathscr{C})$ the constant functor. If you want to be precise about this (for instance, if you worry about naturality), you can look at the anima of natural transformations $H\colon\mathsf{B}\times[1]\to\mathscr{C}$ for which $H(0,-)$ equals $\mathrm{Ran}_j X$ and for which $H(-,1)$ equals a diagram which is equivalent to a diagram equivalent to one in which all objects are $Y$ and all morphisms are an identity morphism on $Y$ (and we require the higher morphisms of this anima to respect these boundary conditions). This anima is equivalent to the anima of functors $\mathsf{B}\star[0]\to\mathscr{C}$ which map $\mathsf{B}$ to $\mathrm{Ran}_j X$ and map the cone point to $Y$ (and for which the higher morphisms respect these boundary condition), via standard arguments. The natural zigzag of maps comparing these two functor anima (induced by a zigzag of maps of diagram shapes, hence why it is natural) witnesses this equivalence. But, the latter subanima of functors $\mathsf{B}\star[0]\to\mathscr{C}$ that we considered is exactly the anima of diagrams we saw above.
So, we have found that this anima is naturally equivalent to $\mathscr{C}^\mathsf{B}(\mathrm{Ran}_j X, \Delta Y)$. This in turn is by the adjunction $\mathrm{colim}\dashv\Delta$ naturally equivalent to the anima $\mathscr{C}(\mathrm{colim}\,\mathrm{Ran}_j X, Y)=\mathscr{C}(FX,Y)$. Therefore, we see that $F$ is indeed left adjoint to $\mathrm{Lan}_{i_3}$.
This means that $\Omega$ is right adjoint to the composition $$ \mathscr{C}\xrightarrow{\Delta}\mathrm{Fun}(\mathsf{Sq},\mathscr{C})\xrightarrow{i^*}\mathrm{Fun}(\mathsf{A},\mathscr{C})\xrightarrow{F}\mathscr{C} $$ which we claim is the suspension functor. This boils down to showing that, for any $X\in\mathscr{C}$, the suspension $\Sigma X$ is the colimit of the diagram
The colimit of this diagram may be computed in a natural way by successively taking pushouts as indicated below:
But this makes it clear that this colimit is naturally equivalent to $\Sigma X$. Hence $\Sigma\dashv\Omega$.
Now, what about the adjunction unit? The adjunction unit of a composite of adjunctions is an approrpriate composite of adjunction units. It is not hard to compute that the adjunction unit of $F\dashv\mathrm{Lan}_{i_3}$ is given for $X\colon\mathsf{A}\to\mathscr{C}$ by the map
The adjunction unit of $i^*\dashv\mathrm{Ran}_i$ is given for $Y\colon\mathsf{B}\to\mathscr{C}$ by the map
For the adjunction $\mathrm{Lan}_{i_0}\dashv i_0^*$ and $Y\in\mathscr{C}$, the adjunction unit is simply $\mathrm{id}_Y$. So, starting with $X\in\mathscr{C}$, the adjunction unit of $\Sigma\dashv\Omega$ equals the composition $$ X\xrightarrow{\mathrm{id}} X\xrightarrow{\mathrm{id}\times_{\mathrm{id}}\mathrm{id}} {X\times_X X} \xrightarrow{p_1\times_{p}p_2}{*\times_{\Sigma X}*\simeq\Omega\Sigma X} $$ where $p\colon X\to \Sigma X$ is the canonical map coming from the pushout square. But this is precisely the map $\eta\colon X\to \Omega\Sigma X$ that you were after.