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1. • CommentRowNumber1.
   • CommentAuthorHarry Gindi
   • CommentTimeFeb 8th 2011
   • (edited Feb 8th 2011)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexThe statement of HTT Lemma 3.2.2.9 refers to "the map $\Delta^n\times A^n \to M(\Phi)$", but unfortunately, it does not actually define what this map is or what properties we expect it to have (and it is not defined anywhere earlier in the book). Is there some canonical choice of such a map that I'm somehow missing? Got it, $A^n\times \Delta^n$ is the constant mapping simplex of length $n$ at the simplicial set $A^n$. The map is induced by the functoriality of the mapping simplex.

   The statement of HTT Lemma 3.2.2.9 refers to “the map $\Delta^n\times A^n \to M(\Phi)$”, but unfortunately, it does not actually define what this map is or what properties we expect it to have (and it is not defined anywhere earlier in the book). Is there some canonical choice of such a map that I’m somehow missing?

   Got it, $A^n\times \Delta^n$ is the constant mapping simplex of length $n$ at the simplicial set $A^n$. The map is induced by the functoriality of the mapping simplex.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeFeb 8th 2011
   • (edited Feb 8th 2011)
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   Author: Urs
   Format: MarkdownItex> Is there some canonical choice of such a map Yes. Maybe you want to explicitly write $$ \Delta^n \times A^n \simeq \coprod_{\sigma \in A^n} \Delta^n $$ which in turn you should think of as $$ \cdots = \coprod_{\sigma : \Delta^n \to A} \Delta^n \,. $$ Then go back to the definition of $M(\phi)$ on the top of p. 151: a morphism $\Delta^n \to M(\phi)$ is a pair consisting of a morphism $f : \Delta^n \to \Delta^n$ and a morphism $\sigma : \Delta^n \to A^{max f}$. The first has a canonical choice: the identity. The second choice is parameterized by the index set $A^n$ of the above coproducts.

   Is there some canonical choice of such a map

   Yes. Maybe you want to explicitly write

   $$\Delta^n \times A^n \simeq \coprod_{\sigma \in A^n} \Delta^n$$

   which in turn you should think of as

   $$\cdots = \coprod_{\sigma : \Delta^n \to A} \Delta^n \,.$$

   Then go back to the definition of $M(\phi)$ on the top of p. 151: a morphism $\Delta^n \to M(\phi)$ is a pair consisting of a morphism $f : \Delta^n \to \Delta^n$ and a morphism $\sigma : \Delta^n \to A^{max f}$.

   The first has a canonical choice: the identity. The second choice is parameterized by the index set $A^n$ of the above coproducts.

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeFeb 8th 2011
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   Author: Urs
   Format: MarkdownItexWait a sec, I didn't use the indices as in the book. Just a sec

   Wait a sec, I didn’t use the indices as in the book. Just a sec

4. • CommentRowNumber4.
   • CommentAuthorHarry Gindi
   • CommentTimeFeb 8th 2011
   • (edited Feb 8th 2011)
   • PermaLink
   Author: Harry Gindi
   Format: MarkdownItexUrs, it's even easier than that. Given a simplicial set $A$, let $\c_A:[n]\to sSet$ be the constant functor at $A$. Claim: $M(c_A)\cong A\times \Delta^n$. Giving a simplex of $\Delta^j\to M(c_A)$ gives a map $\Delta^j\to \Delta^n$ and another map $\Delta^j\to \A$ (since $c_A$ is constant). That is the same thing as giving a simplex of the product, by definition, so we're done. Then we obtain the required map (the one I originally wanted) by the functoriality of the mapping simplex functor, since we have an obvious natural transformation $c_{A^n}\to \Phi$ defined component-wise as follows: $\alpha_i:A^n\to A^i$ is given by the composite $A^n \to A^{n-1}\to\dots\to A^i$. We immediately verify that the transformation is natural (by construction).

   Urs, it’s even easier than that.

   Given a simplicial set $A$, let $\c_A:[n]\to sSet$ be the constant functor at $A$.

   Claim: $M(c_A)\cong A\times \Delta^n$.

   Giving a simplex of $\Delta^j\to M(c_A)$ gives a map $\Delta^j\to \Delta^n$ and another map $\Delta^j\to \A$ (since $c_A$ is constant). That is the same thing as giving a simplex of the product, by definition, so we’re done.

   Then we obtain the required map (the one I originally wanted) by the functoriality of the mapping simplex functor, since we have an obvious natural transformation $c_{A^n}\to \Phi$ defined component-wise as follows:

   $\alpha_i:A^n\to A^i$ is given by the composite $A^n \to A^{n-1}\to\dots\to A^i$. We immediately verify that the transformation is natural (by construction).

5. • CommentRowNumber5.
   • CommentAuthorHarry Gindi
   • CommentTimeFeb 9th 2011
   • (edited Feb 9th 2011)
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   Author: Harry Gindi
   Format: MarkdownItexHmm, a more pressing concern: How can we show that $M(\Phi)\cong M(\Phi')\coprod_{A^n\times \Delta^{n-1}}A^n\times \Delta^n$? It's a pain in the neck because that pushout is defined by a mapping property in, while the pushout has a mapping property in the opposite direction. Here's a [link](http://mathoverflow.net/questions/54866/the-pushout-of-a-mapping-simplex-along-products-is-a-mapping-simplex) to the mathoverflow question.

   Hmm, a more pressing concern:

   How can we show that $M(\Phi)\cong M(\Phi')\coprod_{A^n\times \Delta^{n-1}}A^n\times \Delta^n$?

   It’s a pain in the neck because that pushout is defined by a mapping property in, while the pushout has a mapping property in the opposite direction.

   Here’s a link to the mathoverflow question.