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In Lemma 5.4.5.11 of HTT, the proof given relies on Lemma 5.4.5.10. However it seems that Lurie applies Lemma 5.4.5.10, which requires the given simplicial set to be contractable, to an arbitrary $\kappa$-small simplicial set.

This seeming incongruity was pointed out in this question on mathoverflow 2 years ago. However an answer was never given, and therefore I thought I might re-ask this in a new question (Let me know if there is a better way to re-ask an unanswered question).

Is there either
a) a way to salvage the proof given, or
b) a new proof which avoids the issue, or
c) is the proof actually correct (and we're all being daft)

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  • $\begingroup$ I think (b). Try proving the lemma by hand in the case of 1-categories and when K consists of a single object and K' of two objects (so that we cannot apply 5.4.5.10). At the point where the book uses 5.4.5.10, you'll see that you still need to use the fact that J is filtered and then that f is cofinal (in order to end up with a diagram in I). That should give you an idea of how the general proof should go, though I haven't worked out the details. $\endgroup$ Commented Mar 10, 2021 at 17:00
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    $\begingroup$ @ReidBarton I tried something along the lines of this, but in fact I couldn't see how we could apply the fact that f is $\kappa$-cofinal (note at this point that we in fact do not know that $f$ is cofinal, that is proven next as a consequence of this assertion). In particular being $\kappa$-cofinal only allows you to find weak pre-images for slice categories of contractable simplicial sets. In this situation we are sliced under $K$, which is not assumed to be weakly contractable. $\endgroup$ Commented Mar 11, 2021 at 9:22

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I think there is a typo in Lemma 5.4.5.11: $K$ is supposed to be $\tau$-small and not $\kappa$-small. Note that if $\tau < \kappa$ and $K$ is $\kappa$-small but not $\tau$-small then the statement of the lemma is simply false: e.g., set $\mathcal{I}=\mathcal{J}=K=\mathbb{N}$ to be the poset of natural numbers (with arrows pointing from small numbers to larger numbers) and take both $f$ and $p$ to be the identity. Set $\tau=\omega$ and $\kappa$ an uncountable cardinal. Then $\mathcal{I}_{p/}$ is empty, and so certainly not $\tau$-filtered.

Note that the second statement in 5.4.5.11 is independent of this problem, and its proof does not use any size bound on $K$ (there is a small typo though in the first line of the proof of (2), "where $K$ is now $\kappa$-small and weakly contractible" -> $K$ should be $K'$.

Note also that 5.4.5.11 is cited two times in HTT, once in 5.4.5.12 and once in 5.4.6.5, but in both cases $K=\Delta^0$ and is in particular $\tau$-small.

The issue will hence in principle be resolved if we prove 5.4.5.11 under this modified assumption:

Proof of 5.4.5.11(1) when $K$ is $\tau$-small

Let $\tilde{q}:K' \to \mathcal{I}_{p/}$ be a $\tau$-small diagram classifying a compatible pair of maps $q: K' \to \mathcal{I}$ and $q':K \star K' \to \mathcal{J}$. Since $\mathcal{I}$ is $\tau$-filtered we can find an extension $\overline{q}:(K')^{\triangleright} \to \mathcal{I}$ of $q$. To facilitate notation later on let us write $L:= (K')^{\triangleright}$ and let $l \in L$ be the vertex corresponding to the cone point of $(K')^{\triangleright}$. Now $q'$ and $f\overline{q}$ combine to give a map $r:[K \star K'] \coprod_{K'}L \to \mathcal{J}$. Since $\mathcal{J}$ is $\tau$-filtered and $K, L$ are $\tau$-small we can find an extension of $r$ to a map $$ \overline{r}:\Big[[K \star K']\coprod_{K'} L \Big]^{\triangleright} \to \mathcal{J} .$$ Let $x := \overline{q}(l)$ and $\alpha: f(x) \to y$ be the arrow corresponding to the restriction of $\overline{r}$ to $\Delta^1 = \{l\}^{\triangleright} \subseteq L^{\triangleright}$. Since $f$ is $\kappa$-cofinal there exists an arrow $\beta: x \to z$ in $\mathcal{I}$ and a map $\eta:\alpha \to f(\beta)$ in $\mathcal{J}_{f(x)/}$. Since the inclusion $\{l\} \subseteq L$ is right anodyne we have that $L\coprod_{\{l\}} [\{l\} \star \Delta^0] \subseteq L \star \Delta^0$ is inner anodyne and so may now extend the map $L\coprod_{\{l\}} [\{l\} \star \Delta^0] \to \mathcal{I}$ determined by $\overline{q}$ and $\beta$ to a map $\overline{q}_{\beta}:L \star \Delta^0 \to \mathcal{I}$. Next since $L \coprod_{\{l\}} \{l\}^{\triangleright} \subseteq L^{\triangleright} \subseteq \Big[[K \star K']\coprod_{K'} L \Big]^{\triangleright} $ is a sequence of an inner anodyne map followed by a right anodyne map we may extend the map $$ \Big[[K \star K']\coprod_{K'} L \Big]^{\triangleright} \coprod_{L} [L \star \Delta^0] \coprod_{\{l\} \star \Delta^0} [\{l\}^{\triangleright} \star \Delta^0] = $$ $$ \Big[[K \star K']\coprod_{K'} L \Big]^{\triangleright} \coprod_{\big[L \coprod_{\{l\}} \{l\}^{\triangleright}\big]} \Big[\big[L \coprod_{\{l\}} \{l\}^{\triangleright}\big] \star \Delta^0\Big]\to \mathcal{J} $$ determined by $\overline{r}$, $f\overline{q}_{\beta}$ and $\eta$ to a map $$ \overline{r}_{\eta}:\Big[[K \star K']\coprod_{K'} L \Big]^{\triangleright} \star \Delta^0 \to \mathcal{J} .$$ The restriction of $\overline{r}_{\eta}$ to $K \star K'\star \Delta^0$ and the restriction of $\overline{q}_{\beta}$ to $K'\star \Delta^0$ now determine an extension of $\tilde{q}:K' \to \mathcal{I}_{p/}$ to a map $K'\star \Delta^0 \to \mathcal{I}_{p/}$, as needed.

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  • $\begingroup$ Just went through the proof, looks completely correct to me. So does the counterexample. Thanks a lot for that! $\endgroup$ Commented Mar 11, 2021 at 11:39

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