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\author{Gabriel C. Drummond-Cole}
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\title{Deformation Theory and Operads}
\begin{document}\maketitle
Let me give the simplicial interpretation of the Hochschild complex.  Let's start with a prelude on simplicial objects. 
\begin{defi}Let $\simp$ be the category with objects $\bar{n}=\{0,\ldots, n\}$ for $n\ge 0$ and morphisms ar maps $f:\{0,\ldots,n\}\to\{0,\ldots,n\}$, non-decreasing.  Composition is composition of set maps.
\end{defi}
The most important maps are the maps $\delta_i:\bar{n}\to\bar{n+1}$ which skips $i$ and is injective.  Let $\sigma_i:\bar{n}\to\bar{n-1}$ be the surjective map that hits $i$ twice.

As homework, every non-decreasing map $f:\bar{n}\to\bar{m}$ can be written as $f=\delta_{i_1}\circ\cdots \delta_{i_r}\circ \sigma_{j_1}\circ\cdots\sigma_{j_s}$.

This is unique if $i_r\le \cdots i_1$ and $j_s\le \cdots\le j_1$.  We have the relations $\delta_j\delta_i=\delta_i\delta_{j-1}$ for $i<j$ and $\sigma_j\sigma_i=\sigma_i\sigma_{j+1}$ for $i\le j$ along with $\sigma_j\delta_i=\delta_i\sigma_{j-1}$ for $i<j$, $id_{\bar{n}}$ for $i=j$ or $i=j+1$, and $\delta_{i-1}\sigma_j$ for $i>j+1$

\begin{defi}
Let $\Cc$ be a category.  A cosimplicial object in $\Cc$ is a covariant functor $X^.\simp\to \Cc$
\\
A simplicial object in $\Cc$ is a contravariant functor $X_.:\simp\to \Cc$
\end{defi}
\begin{lemma}
\begin{itemize}
\item[a]A cosimplicial object in $\Cc$ is equivalent to a sequence of objects $X^i\in Obj(\Cc)$ and morphisms in $\Cc$ $D_i:X^n\to X^{n+1}$ for $0\le i\le n+1$, $S_i:X^n\to X^{n-1}$ for $0\le i\le n-1$ satisfying the corresponding commutation relations.
\item[b]A simplicial object is a sequence of objects $X_0,X_1,\ldots$ and morphisms $d_i:X_{n+1}\to X_n$ for $0\le i\le n+1$ ando $s_i:X_{n-1}\to X_n$ for $0\le i\le n-1$
satisfying
$d_id_j=d_{j-1}d_i$ for $i<j$, $s_is_j=s_{j+1}s_i$ for $i\le j$ and then
$$d_is_j=\left\{\begin{array}{lcl}s_{j-1}d_i&,&i<j\\id_{X_n}&,&i=j,i=j+1\\s_jd_{i-1}&,&i>j+1\end{array}\right.$$
\end{itemize}
\end{lemma}
The proof for the first part is, take $X^n=X^.(\bar{n})$ and $D_i=X^.(\delta_i)$,$S_i=X^.(\sigma_i)$.  This works both forward and backward.  The proof of the second is the same.

The most common example is to have $\Cc$ be topological spaces and then let $\Delta^n=\{(t_1,\ldots, t_n)\}|0\le t_1\le \cdots \le 1$

This can be made into a cosimplicial topological space: $\Delta^.:\simp\to Top$ takes $\Delta^n$ as above and $D_i:\Delta^n\to \Delta^{n+1}$ which maps $t_1\,\ldots, t_n$ to $t_1,\ldots, t_i,t_i,\ldots,t_n$ and $S_i$ which maps $t_1,\ldots, t_n\mapsto t_1\,ldots, \hat{t}_i,\ldots, t_n$

Let $X$ be a topological space.  It has no bullet so it's not simplicial or cosimplicial.  The singular $n$-simplices of $X$ are $S(X)_n:=\{r:\Delta^n\to X|r\text{ continuous}\}$.  These can be put together to give a simplicial set.

This cosimplicial thing can be composed with the contravariant functor $Hom(\bullet, X)$ to give this as a contravariant functor $\simp\to Set$.  This says $d_i(r)=r\circ D_i$ and $s_i(r)=r\circ S_i$.  So $S(X)$ is a functor from topological spaces to simplicial sets.

I want to say what the geometric realization is, which is a functor in the opposite direction.  It comes up all over the place.  We have a geometric realization functor $T=|\quad|$.  This takes, for a simplicial set $X_.$ it takes 
$$T(X_.)=|X_.|=\amalg_{n\ge 0} X_n\times \Delta^n/\sim$$
Where the equivalence relation says $(d_i(x),t_1,\ldots, t_n)\sim (x,D_i(t_1,\ldots, t_n))$ and $(s_i(x),t_1,\ldots, t_n)\sim (x,S_i(t_1,\ldots, t_n))$.

These are adjoint functors so there is a bijection between $Hom_{Top}(|X_.|,Y)\to Hom_{Simp\ Set}(X_.,S(Y)_.)$ which is almost tautological.  I start with  $f:\amalg X_n\times \Delta^n /\sim\to Y$.  I want to give a bunch of maps $f_n:X_n\to Hom(\Delta^n, Y)$.  If you look at this this is almost immediate.  You define $f_n(x)(t_1,\ldots, t_n)$ as $f((x,t_1,\ldots, t_n))$.

If $X_.$ is a simplicial set then $x\in X_n$ is called degenerate if $x=s_i(y)$ for some $y$ and $i$.  Otherwise it is nondegenerate.

I want to do one more different example, a simplicial set that gives the circle.  Next time I'll do more examples.  So the simplicial circle $S^1_.$  The goal is to define $S^1_0,\ldots, S^1_i$ which should give the simplices so that the geometric realization is the circle.  So $S^1_0=\{0\}$.  Then $S^1_1$ has $s_0(0)$ so I need at least one more simplex, one that is nondegenerate.  I can let the set be $\{0,1\}$.  I want $d_is_i=id_{X_n}$.  So $s_i$ must be injective.  I don't know how much I want to go through this, but it turns out the smallest thing you can do is the following:
$$\begin{array}{cc}
S^1_0&0
\\S^1_1&0,1
\\S^1_2&0,1,2
\\S^1_3&0,1,2,3
\\S^1_4&0,1,2,3,4
\end{array}$$
If you analyze the $s_i$ and $d_i$ you get that $s_i:S^1_n\to S^1_{n+1}$ for $0\le i\le n$ is given by $s_i(k)$ is either $k$ for $k\le i$ or $k+1$ for $k>i$.

The $d_i$ are given by $d_i(k)=k$ if $k\le i$ and $k-1$ if $k>i$.  This gives all of them but $d_n(k)$ which is $k$ if $k\le n-1$ or $0$ if $k=n$.  So $d_5$ is not a nondecreasing map.  I'm already fifteen minutes late.  I want to show that the geometric realization has just the two nondegenerate simplices.

\end{document}
