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I have a question about the statement of Lemma 5.5.2.1 in Lurie's `Higher Topos Theory'.

``Let $S$ be a small simplicial set, let $f: S\rightarrow \mathcal{S}$ be an object of $\mathcal{P}(S^{op})$, and let $F: \mathcal{P}(S^{op})\rightarrow \widehat{\mathcal{S}}$ be the functor corepresented by $f$. Then the composition $$S\overset{j}{\rightarrow}\mathcal{P}(S^{op})\overset{F}{\rightarrow}\widehat{\mathcal{S}}$$ is equivalent to $f$. "

The Yoneda embedding $j$ should be from $S$ to $\mathcal{P}(S)$, but the last line doesn't seem to be just a typo. Do I miss something? Thanks!

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    $\begingroup$ The usual Yoneda embedding is from $S^{\operatorname{op}}$ to $\mathcal{P}(S)$ - to get $j$, apply the Yoneda embedding to $S^{\operatorname{op}}.$ $\endgroup$
    – dhy
    Commented Nov 30, 2018 at 3:20
  • $\begingroup$ Thanks for fixing HTT. That is my reason for the vote up. $\endgroup$
    – Piotr Hajlasz
    Commented Nov 30, 2018 at 3:55
  • $\begingroup$ By the way, we field short questions like this in the homotopy theory chat here. $\endgroup$
    – Harry Gindi
    Commented Nov 30, 2018 at 4:31
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    $\begingroup$ The second and third P(S^op)’s should have an extra op on the outside, yeah? That ought to do it. $\endgroup$
    – Dylan Wilson
    Commented Nov 30, 2018 at 6:10
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    $\begingroup$ @dhy No, this is actually a typo. The usual Yoneda embedding goes from $S\hookrightarrow \mathcal{P}(S)$. Presheaves are contravariant. Dylan's comment makes this correct. $\endgroup$
    – Harry Gindi
    Commented Nov 30, 2018 at 8:16

1 Answer 1

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The middle term should be $\mathcal{P}(S^{op})^{op}$. The Yoneda embedding gives a functor $S^{op} \to \mathcal{P}(S^{op})$. The functor corepresented by $f \in \mathcal{P}(S^{op})$ is given by $Hom_{\mathcal{P}(S^{op})}(-,f)$, which is a functor $\mathcal{P}(S^{op})^{op} \to \widehat{\mathcal{S}}$.

As a reality check, the analogous theorem in ordinary category is that the composite $$ x \mapsto Hom_S(x,-) \mapsto Nat(Hom_S(x,-),f) $$ is naturally isomorphic via the Yoneda lemma to the functor $x \mapsto f(x)$.

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  • $\begingroup$ Great, thanks for the answer! I was confused about corepresented v.s represented. $\endgroup$
    – Wonderfield
    Commented Nov 30, 2018 at 13:10

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