2
$\begingroup$

In page 27 in HTT of J.Lurie, the expression $$\text{Map}_S(X,Y):=Y^X\times_{S^X}\{\phi\}\in \text{Set}_\Delta$$ appears for simplicial set $X,Y,S$ in Warning 1.2.2.2. However, I couldn't understand two notation in this expression, first one is exponential of simplicial set and the second one is product which have lower index. I guess that $X^Y$ is kind of $Hom_{\text{Set}_\Delta}(X,Y)$, but I couldn't be sure about it.

$\endgroup$
  • 2
    $\begingroup$ $Y^X$ is the simplicial set whose $n$-simplices are the maps $X\times\Delta^n\to Y$, that is the internal hom in simplicial sets. $\endgroup$ – Denis Nardin Jan 13 at 15:45
8
$\begingroup$

Just think in terms of ordinary sets for the moment. We have sets and maps $X\xrightarrow{\phi}S\xleftarrow{\psi}Y$ and we want to think about the set $$ \text{Map}_S(X,Y) = \{f\colon X\to Y: \psi f=\phi\}. $$ We can think of $f$ as an element of $Y^X$, and composition with $\psi$ gives a map $\psi_*\colon Y^X\to S^X$, and the condition $\psi f=\phi$ can be written as $\psi_*(f)=\phi$, so $\text{Map}_S(X,Y)$ is the preimage under $\psi_*$ of the set $\{\phi\}\subset S^X$. In other words, the diagram $\require{AMScd}$ \begin{CD} \text{Map}_S(X,Y) @>>>\{\phi\} \\ @VVV @VVV \\ Y^X @>>> S^X \end{CD} is a pullback. This is the meaning of Lurie's notation. Everything works in essentially the same way for simplicial sets.

$\endgroup$

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.