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1)Let $\mathcal{C}$ be a cartesian closed small category. Let $\operatorname{Map}\: : \: sPsh(\mathcal{C})\times sPsh(\mathcal{C})\to sPsh(\mathcal{C})$ be the internal Hom of simplicial presheaves, i.e for any $C\in \mathcal{C}$ and any $n$ $$ \operatorname{Map}(X,Y)_{n}(C):=\operatorname{Hom}_{sPsh(\mathcal{C})}(X\times y(C)\times \Delta[n], Y), $$ where $y$ is the Yoneda embedding. Assume that the category of simplicial presheaves is equipped with the projective model structure and fix $V\in \mathcal{C}$. Choose a $X\in sPsh(\mathcal{C})$ cofibrant. When $\operatorname{Map}(y(V),X)$ is cofibrant as well?

I asked this because if $X$ is itself representable then $\operatorname{Map}(y(V),X)$ is again cofibrant (it satisfies two conditions of Necessary conditions for cofibrancy in global projective model structure on simplicial presheaves). So I wonder which conditions should satisfy a cofibrant simplicial presheaf $X$ such that $\operatorname{Map}(y(V),X)$ is again cofibrant.

2) Dimitri writes we should assume that $\mathcal{C}$ is cartesian closed. So probably I make a mistake in my computations but I don't see why: assume that $\mathcal{C}$ is the category of manifolds (so NOT cartesian closed), then for $U\in \mathcal{C}$ we have $$ \operatorname{Map}(y(V),y(U))_{n}(C)=\operatorname{Hom}_{sPsh(\mathcal{C})}(y(V)\times y(C)\times \Delta[n], y(U)) $$ which is isomorphic to $\operatorname{Hom}_{\mathcal{C}}(V\times C, U)$. We consider two maps

a) $i\: : \: y(U)\to\operatorname{Map}(y(V),y(U))$ that sends $ f\: : \: C\to U$ to $f\circ \pi_{C}\: : \: C\times V\to C\to U$

b) Fix a $p\in V$. Then let $restr_{p}\: : \: \operatorname{Map}(y(V),y(U))\to y(U)$ be the map that sends $f\: : \: C\times V\to U$ to its restriction at $p\times C$.

Now it seems to me that $i,restr_{p}$ makes $$ y(U)\to\operatorname{Map}(y(V),y(U))\to y(U) $$ a retract of representables(hence it satisfies condition 1 in the link). Where is the mistake?

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  • $\begingroup$ Condition (1) in the link is not satisfied when X is representable, unless C is cartesian closed. For example, it fails when X is representable and C is the category of smooth manifolds. $\endgroup$ – Dmitri Pavlov Jun 24 '15 at 15:12
  • $\begingroup$ You have shown that the representable presheaf is a retract of the mapping space; what you need to show is that the mapping space is a retract of a disjoint union of representable presheaves. To me, that seems unlikely. $\endgroup$ – Zhen Lin Jun 25 '15 at 9:21
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    $\begingroup$ It seems that I forgot what does it means retract! Thanks a lot. Now I see why we need that $\mathcal{C}$ is cartesian closed. Thanks $\endgroup$ – Cepu Jun 25 '15 at 9:32

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