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Mar 1, 2020 at 4:18 comment added Tim Campion One subtlety is that $L \underline X$ is a priori only connected to $X$ by a zigzag of weak equivalences passing outside the Brown-Golasinski category. But that's ok -- $L \underline X$ is cofibrant as an $\infty$-category with strict inverses, and any homotopy equivalence between products of Eilenberg-MacLane spaces lifts to an equivalence of $H \mathbb Z$-modules, so there exists a direct Brown-Golasinski equivalence $L \underline X\to X$. Thus we are indeed justified in computing the derived Brown-Golasinski homotopy classes of maps from $X$ to $Y$ using $L \underline X$.
Mar 1, 2020 at 3:56 comment added Tim Campion Thus we have certain objects in the homotopy category of strict $\infty$-categories with weak inverses -- namely those objects whose inverses are strict -- which have smaller homspaces as strict $\infty$-categories than they have as weak $\infty$-categories. So the inclusion is not fully faithful.
Mar 1, 2020 at 3:54 comment added Tim Campion And homotopies with respect to $I \underline X$ are in bijection with homotopies with respect to $L(I \underline X)$. Now, maps in the homotopy category of strict $\infty$-categories from $X$ to $Y$ can be described as $Hom(\underline X,Y)$ modulo $I \underline X$ homotopy. Whereas maps in the homotopy category of strict $\infty$-groupoids from $X$ to $Y$ can be described as $Hom(L\underline X, Y)$ module $L (I\underline X)$-homotopy. But these are the same set! So it appears to me that at least at the level of homotopy categories, we get the same set of maps!
Mar 1, 2020 at 3:53 comment added Tim Campion @SimonHenry Hang on though -- at the 1-categorical level, strict $\infty$-categories with all strict inverses are reflective in strict $\infty$-categories; let $L$ denote the reflection functor. If $X,Y$ are strict $\infty$-categories with strict inverses and $\underline X$ is a cofibrant replacement of $X$, then any map $\underline X \to Y$ factors through some map $L \underline X \to Y$. And if $I \underline X$ is a cofibrant cylinder on $\underline X$, then because $L$ is left Quillen, we have that $L(I \underline X)$ is a cofibrant cylinder for $L\underline X$.
Mar 1, 2020 at 2:50 comment added Simon Henry Exactly. Or at least, we don't know if we can. The forgetfull functor from the Brown-Golasinski model structure to the Lafont-Metayer-Worytkiewicz model structure is right Quillen, I do not known how this adjunction behave on the corresponding $(\infty,1)$-categories, ( and I have talked about this with enough people to convince me that this was not known). So as far as I know the left adjoint could be homotopy fully faithfull, in which case your argument would apply.
Mar 1, 2020 at 2:35 comment added Tim Campion @SimonHenry Ah -- I completely failed to appreciate the distinction between strict $\infty$-categories with all strict inverses and strict $\infty$-categories with all weak inverses.... If I have things right, the Brown-Golasinski model structure on the former is projectively-induced from the LaFont-Metayer-Worytkiewicz model structure where the latter live. So fibrancy is not an issue, but an $\infty$-category with strict inverses won't typically be cofibrant in the LaFont-Metayer-Worytkiewicz model structure. So we can't even compute the mapping spaces in the way I was assuming...
Mar 1, 2020 at 1:59 comment added Simon Henry ... more maps than in the localization of the category of strict $\infty$-groupoid. As far as I know it is an open question whether the full subcategory of the localization of the folk model structure on "reversible $\infty$-category" is equivalent to the localization of the category of strict $\infty$-groupoids. But of course I definitely agree that the answer is most likely no.
Mar 1, 2020 at 1:56 comment added Simon Henry I'm not completely agreeing with your answer (though I believe you are probably right in the end). My objection is as follows: we need to distinguishes the strict $\infty$-groupoid where every arrow is strongly invertible from these where every arrows is only weekly invertible (I'll call these reversible arrows). When you compute map between two strict $\infty$-groupoids in the folk model structure you need to consider a resolution of the $\infty$-groupoid on the left by an equivalent $\infty$-category, where the arrow will only be reversible and not invertible, so there is potentially...
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