@ -11,7 +11,7 @@ Before defining \emph{simplicial abelian groups}, we will first discuss the more
There are two special kinds of maps in $\DELTA$, the so called \emph{face} maps and \emph{degeneracy} maps. The \emph{$i$-th face maps}$\delta_i: [n-1]\to[n]$ is the unique injective monotone function which \emph{omits}$i$. More precisely, it is defined for all $n \in\Np$ as (note that we do not explicitly denote $n$ in this notation)
$$\delta_i: [n-1]\to[n], k \mapsto\begin{cases} k &\text{if } k < i,\\ k+1&\text{if } k \geq i, \end{cases}\hspace{1.0cm}0\leq i \leq n. $$
The \emph{$i$-th degeneracy map}$\sigma_i: [n+1]\to[n]$ is the unique surjective monotone function which \emph{hits $i$ twice}. More precisely it is defined for all $n \in\N$ as:
The \emph{$i$-th degeneracy map}$\sigma_i: [n+1]\to[n]$ is the unique surjective monotone function which \emph{hits $i$ twice}. More precisely it is defined for all $n \in\N$ as
$$\sigma_i: [n+1]\to[n], k \mapsto\begin{cases} k &\text{if } k \leq i,\\ k-1&\text{if } k > i, \end{cases}\hspace{1.0cm}0\leq i \leq n. $$
The nice things about these maps is that every map in $\DELTA$ can be decomposed to a composition of such maps. So in a sense, these are all the maps we need to consider.
@ -96,7 +96,7 @@ Of course the maps $\delta_i$ and $\sigma_i$ in $\DELTA$ satisfy certain relatio
Note that these cosimplicial identities are ``purely categorical'', i.e. they only use compositions and identity maps. Because a simplicial set $X$ is a contravariant functor, dual versions of these equations hold in its image. For example, the first equation corresponds to $X(\delta_i)X(\delta_j)= X(\delta_{j-1})X(\delta_i)$ for $i < j$. This can be used for an explicit definition of simplicial sets. In this definition a simplicial set $X$ consists of a collection of sets $X_n$ together with face and degeneracy maps. More precisely:
\begin{definition}
\emph{(Explicitly)} An simplicial set $X$ consists of a collection sets $X_n$ together with functions $d_i: X_n \to X_{n-1}$ and $s_i: X_n \to X_{n+1}$ for $0\leq i \leq n$ and $n \in\N$, such that the simplicial identities hold
\emph{(Explicitly)} An simplicial set $X$ consists of a collection sets $X_n$ together with functions $d_i: X_n \to X_{n-1}$ and $s_i: X_n \to X_{n+1}$ for $0\leq i \leq n$ and $n \in\N$, such that the simplicial identities hold
\begin{align}
d_i d_j &= d_{j-1} d_i, \hspace{1.5cm}\text{ if } i < j,\\
d_i s_j &= s_{j-1} d_i, \hspace{1.5cm}\text{ if } i < j,\\
@ -106,30 +106,30 @@ Note that these cosimplicial identities are ``purely categorical'', i.e. they on
\end{align}
\end{definition}
It is already indicated that a functor from $\DELTA^{op}$ to $\Set$ is determined when the images for the face and degeneracy maps in $\DELTA$ are provided. So this gives a way of restoring the first definition from this one. Conversely, we can apply functoriality to obtain the second definition from the first. So these definitions are the same. From now on we will denote $X([n])$ by $X_n$, $X(\sigma_i)$ by $s_i$ and $X(\delta_i)$ by $d_i$, whenever we have a simplicial set $X$. For any other map $\beta : [n]\to[p]$ we will denote the induced map by $\beta^\ast: X_p \to X_n$.
It is already indicated that a functor from $\DELTA^{op}$ to $\Set$ is determined when the images for the face and degeneracy maps in $\DELTA$ are provided. So this gives a way of restoring the first definition from this one. Conversely, we can apply functoriality to obtain the second definition from the first. So these definitions are the same. From now on we will denote $X([n])$ by $X_n$, $X(\sigma_i)$ by $s_i$ and $X(\delta_i)$ by $d_i$, whenever we have a simplicial set $X$. For any other map $\beta : [n]\to[p]$ we will denote the induced map by $\beta^\ast: X_p \to X_n$.
When using a simplicial set to construct another object, it is often handy to use this second definition, as it gives you a very concrete objects to work with. On the other hand, constructing this might be hard (as you would need to provide a lot of details), in this case we will often use the more abstract definition.
Note that because of the third equation, the degeneracy maps $s_i$ are injective. This means that in the set $X_{n+1}$ there are always ``copies'' of elements of $X_n$. In a way these elements are not interesting, hence we call them degenerate.
\begin{definition}
An element $x \in X_{n+1}$ is \emph{degenerate} if it lies in the image of $s_i: X_n \to X_{n+1}$ for some $i$, otherwise it is called \emph{non-degenerate}.
An element $x \in X_{n+1}$ is \emph{degenerate} if it lies in the image of $s_i: X_n \to X_{n+1}$ for some $i$, otherwise it is called \emph{non-degenerate}.
\end{definition}
\begin{lemma}
\label{le:non-degenerate}
We can write any $x \in X_n$ uniquely as $x =\beta^\ast y$ with $\beta: [n]\epi[m]$ a surjective map and $y \in X_m$ non-degenerate.
We can write any $x \in X_n$ uniquely as $x =\beta^\ast y$ with $\beta: [n]\epi[m]$ a surjective map and $y \in X_m$ non-degenerate.
\end{lemma}
\begin{proof}
We will proof the existence by induction over $n$. For $n=0$ the statement is trivial, since all elements in $X_0$ are non-degenerate. Assume the statement is proven for $n$. Let $x \in X_{n+1}$. Clearly if $x$ itself is non-degenerate, we can write $x =\id^\ast x$. Otherwise it is of the form $x = s_i x'$ for some $x' \in X_n$ and $i$. The induction hypothesis tells us that we can write $x' =\beta^\ast y$ for some surjection $\beta: [n]\epi[m]$ and $y \in X_m$ non-degenerate. So $x = s_i \beta^\ast y =(\beta\sigma_i)^\ast y$.
For uniqueness, assume $x =\beta^\ast y =\gamma^\ast z$ with $\beta: [n]\epi[m]$, $\gamma: [n]\epi[m']$ and $y \in X_m, z \in X_{m'}$ non-degenerate. Because $\beta$ is surjective there is an $\alpha:[m]\to[n]$ such that $\beta\alpha=\id$ and hence $y =\alpha^\ast\gamma^\ast z =(\gamma\alpha)^\ast z$. By the epi-mon factorization (Lemma~\ref{le:epimono}) we can write $\gamma\alpha=\delta_{i_a}\cdots\delta_{i_1}\sigma_{j_b}\cdots\sigma_{j_1}$, using that $y$ is non-degenerate we know that $\gamma\alpha$ is injective. So we have $\gamma\alpha: [m]\mono[m']$. Because of symmetry (of $y$ and $z$) we also have some map $[m']\mono[m]$, so $m = m'$. So $\gamma\alpha$ is also surjective, hence the identity function, thus $y = z$.
For uniqueness, assume $x =\beta^\ast y =\gamma^\ast z$ with $\beta: [n]\epi[m]$, $\gamma: [n]\epi[m']$ and $y \in X_m, z \in X_{m'}$ non-degenerate. Because $\beta$ is surjective there is an $\alpha:[m]\to[n]$ such that $\beta\alpha=\id$ and hence $y =\alpha^\ast\gamma^\ast z =(\gamma\alpha)^\ast z$. By the epi-mon factorization (Lemma~\ref{le:epimono}) we can write $\gamma\alpha=\delta_{i_a}\cdots\delta_{i_1}\sigma_{j_b}\cdots\sigma_{j_1}$, using that $y$ is non-degenerate we know that $\gamma\alpha$ is injective. So we have $\gamma\alpha: [m]\mono[m']$. Because of symmetry (of $y$ and $z$) we also have some map $[m']\mono[m]$, so $m = m'$. So $\gamma\alpha$ is also surjective, hence the identity function, thus $y = z$.
Now assume $x =\beta^\ast y =\gamma^\ast y$ with $\gamma, \beta: [n]\epi[m]$ such that $\beta\neq\gamma$, and $y \in X_m$ non-degenerate. Then we can find an $\alpha:[m]\to[n]$ such that $\beta\alpha=\id$ and $\gamma\alpha\neq\id$. With the epi-mono factorization write $\gamma\alpha=\delta_{i_a}\cdots\delta_{i_1}\sigma_{j_b}\cdots\sigma_{j_1}$, then by functoriality of $X$
Now assume $x =\beta^\ast y =\gamma^\ast y$ with $\gamma, \beta: [n]\epi[m]$ such that $\beta\neq\gamma$, and $y \in X_m$ non-degenerate. Then we can find an $\alpha:[m]\to[n]$ such that $\beta\alpha=\id$ and $\gamma\alpha\neq\id$. With the epi-mono factorization write $\gamma\alpha=\delta_{i_a}\cdots\delta_{i_1}\sigma_{j_b}\cdots\sigma_{j_1}$, then by functoriality of $X$
$$ y =\alpha^\ast\beta^\ast y =\alpha^\ast\gamma^\ast y = s_{j_1}\cdots s_{j_b} d_{i_1}\cdots d_{i_a} y. $$
Note that $y$ was non-degenerate, so $s_{j_1}\cdots s_{j_b}=\id$, hence $d_{i_1}\cdots d_{i_a}=\id$. So $\gamma\alpha=\id$, which gives a contradiction. So $\beta$ is unique.
\end{proof}
\subsection{The standard $n$-simplex}
Recall that for any category $\cat{C}$ we have the $\mathbf{Hom}$-functor $\Hom{\cat{C}}{-}{-}: \cat{C}^{op}\times\cat{C}\to\Set$. We can fix an object $C \in\cat{C}$ and get a functor $\Hom{\cat{C}}{-}{C} : \cat{C}^{op}\to\Set$. In our case we can get the following simplicial sets in this way:
Recall that for any category $\cat{C}$ we have the $\mathbf{Hom}$-functor $\Hom{\cat{C}}{-}{-}: \cat{C}^{op}\times\cat{C}\to\Set$. We can fix an object $C \in\cat{C}$ and get a functor $\Hom{\cat{C}}{-}{C} : \cat{C}^{op}\to\Set$. In our case we can get the following simplicial sets in this way:
\begin{definition}
The \emph{standard $n$-simplex}$\Delta[n]\in\sSet$ is given by
@ -139,7 +139,7 @@ Recall that for any category $\cat{C}$ we have the $\mathbf{Hom}$-functor $\Hom{
Note that $\Delta[-]: \DELTA\to\sSet$ is exactly the Yoneda embedding. In a moment we will see why the Yoneda lemma is useful to us, but let us first explicitly describe two examples of such standard simplices.
\begin{example}
We will compute how $\Delta[0]$ look like. Note that $[0]$ is an one-element set, so for any set $S$, there is only one function $\ast: S \to[0]$. Hence $\Delta[0]_n =\{\ast\}$ for all $n$ and the face and degeneracy maps are necessarily the identity maps $\id: \{\ast\}\to\{\ast\}$. Thus, $\Delta[0]$ looks like:
We will compute how $\Delta[0]$ look like. Note that $[0]$ is an one-element set, so for any set $S$, there is only one function $\ast: S \to[0]$. Hence $\Delta[0]_n =\{\ast\}$ for all $n$ and the face and degeneracy maps are necessarily the identity maps $\id: \{\ast\}\to\{\ast\}$. Thus, $\Delta[0]$ looks like
$$\Delta[0] :=
\begin{tikzpicture}[baseline=-0.5ex]
\matrix (m) [matrix of math nodes] {
@ -163,7 +163,7 @@ Note that $\Delta[-]: \DELTA \to \sSet$ is exactly the Yoneda embedding. In a mo
For $\Delta[1]_0$ we have to consider maps from $[0]$ to $[1]$, we cannot first apply degeneracy maps (there is no object $[-1]$). So this leaves us with the face maps: $\Delta[1]_0=\{\delta_0, \delta_1\}$. For $\Delta[1]_1$ we of course have the identity function and two functions $\delta_0\sigma_0, \delta_1\sigma_0$. Now $\Delta[1]_2$ are the maps from $[2]$ to $[1]$.
We will compute the two face maps $d_0$ and $d_1$ from $\Delta[1]_1$ to $\Delta[1]_0$. Recall that the $\mathbf{Hom}$-functor in the first argument (the contravariant argument) works with precomposition. So this gives:
We will compute the two face maps $d_0$ and $d_1$ from $\Delta[1]_1$ to $\Delta[1]_0$. Recall that the $\mathbf{Hom}$-functor in the first argument (the contravariant argument) works with precomposition. So this gives
@ -193,13 +193,13 @@ Note that $\Delta[-]: \DELTA \to \sSet$ is exactly the Yoneda embedding. In a mo
\end{example}
\subsection{Simplicial objects in arbitrary categories}
Of course the definition of simplicial set can easily be generalized to other categories. For any category $\cat{C}$ we can consider the functor category $\cat{sC}=\cat{C}^{\DELTA^{op}}$. In this thesis we are interested in the category of simplicial abelian groups:
Of course the definition of simplicial set can easily be generalized to other categories. For any category $\cat{C}$ we can consider the functor category $\cat{sC}=\cat{C}^{\DELTA^{op}}$. In this thesis we are interested in the category of \emph{simplicial abelian groups}:
$$\sAb=\Ab^{\DELTA^{op}}. $$
So a simplicial abelian group $A$ is a collection of abelian groups $A_n$, together with face and degeneracy maps, which in this case means group homomorphisms $d_i$ and $s_i$ such that the simplicial equations hold.
Note that the set of natural transformations between two simplicial abelian groups $A$ and $B$ is also an abelian group. The proof that $\sAb$ is a preadditive category is very similar to the proof we saw in Section~\ref{sec:Chain Complexes}. For two natural transformations $f,g: A \to B$ we simply define $f+g$ pointwise by $(f+g)_n = f_n + g_n$ and it is easily checked that this is a natural transformation.
As we are interested in simplicial abelian groups, it would be nice to make these standard $n$-simplices into simplicial abelian groups. We have seen how to make an abelian group out of any set using the free abelian group functor. We can use this functor $\Z[-]: \Set\to\Ab$ to induce a functor $\Z^\ast[-]: \sSet\to\sAb$ as shown in the following diagram.
As we are interested in simplicial abelian groups, it would be nice to obtain simplicial abelian groups associated to the standard $n$-simplices. We have seen how to make an abelian group out of any set using the free abelian group functor. We can use this functor $\Z[-]: \Set\to\Ab$ to induce a functor $\Z^\ast[-]: \sSet\to\sAb$ as shown in the following diagram.
The only thing that we need to check is that this bijection preserves the group structure. Recall that this bijection from $\Hom{\sAb}{\Z[\Delta[n]]}{A}$ to $A_n$ is given by (where $\id=\id_{[n]}$ is a generator in $\Z[\Delta[n]]$):
The only thing that we need to check is that this bijection preserves the group structure. Recall that this bijection from $\Hom{\sAb}{\Z[\Delta[n]]}{A}$ to $A_n$ is given by (where $\id=\id_{[n]}$ is a generator in $\Z[\Delta[n]]$)
$$\phi(f)= f_n(\id)\in X_n \quad\text{ for } f: \Delta[n]\to X. $$
Now let $A$ be a simplicial abelian group and $f, g: \Z\Delta[n]\to A$ maps. Then we compute:
Now let $A$ be a simplicial abelian group and $f, g: \Z\Delta[n]\to A$ maps. Then we compute
where we regard $\id\in\Delta[n]$ as an element $\id\in\Z\Delta[n]$, we can do so by the unit of the adjunction. So this bijection is also a group homomorphism, hence we have an isomorphism $\Hom{\sAb}{\Z[\Delta[n]]}{A}\iso A_n$ of abelian groups.
Comparing chain complexes and simplicial abelian groups, one sees a certain similarity. Both concepts are defined as sequences of abelian groups with certain structure maps. At first sight simplicial abelian groups seem to have a richer structure. There are many face maps as opposed to only a single boundary homomorphism. Nevertheless, as we will show in this section, these two concepts give rise to equivalent categories.
\subsection{Unnormalized chain complex}
Given a simplicial abelian group $A$, we have a family of abelian groups $A_n$. For every $n>0$ we define a group homomorphism$\del_n : A_n \to A_{n-1}$
$$\del_n = d_0- d_1+\ldots+(-1)^n d_n.$$
Given a simplicial abelian group $A$, we have a family of abelian groups $A_n$. For every $n>0$ we define a group homomorphism
In this calculation we did the following. We split the inner sum in two halves \refeqn{1} and we use the simplicial equations on the second sum \refeqn{2}. Then we do a shift of indices \refeqn{3}. By interchanging the roles of $i$ and $j$ in the second sum, we have two equal sums which cancel out. So indeed this is a chain complex.
\end{proof}
This construction defines a functor $M : \sAb\to\Ch{\Ab}$. And in fact we already used it in the construction of the singular chain complex, where we defined the boundary maps (on generators) as $\del(\sigma)=\sigma\circ d_0-\sigma\circ d_1+\ldots+(-1)^{n+1}\sigma\circ d_{n+1}$. We will briefly come back to this in Section~\ref{sec:Homotopy}.
Thus, associated to a simplicial abelian group $A$ we obtain a chain complex $M(A)$ with $M(A)_n = A_n$ and the boundary operator as above. This construction defines a functor
$$ M: \sAb\to\Ch{\Ab}$$
by assigning $M(f)_n = f_n$ for a natural transformation $f: A \to B$. It follows from a nice calculation that $M(f)$ is indeed a chain map:
where we used naturality of $f$ in step \refeqn{1}. This functor is in fact already used in the construction of the singular chain complex, where we defined the boundary maps (on generators) as $\del(\sigma)=\sigma\circ d_0-\sigma\circ d_1+\ldots+(-1)^{n+1}\sigma\circ d_{n+1}$. We will briefly come back to this in Section~\ref{sec:Homotopy}.
Let us investigate whether this functor can be used for our sought equivalence. For a functor from $\Ch{\Ab}$ to $\sAb$ we cannot simply take the same collection of abelian groups. This is due to the fact that the degeneracy maps should be injective. This means that for a simplicial abelian group $A$, if we know $A_n$ is non-trivial, then all $A_m$ for $m > n$ are also non-trivial.
Let us investigate whether this functor $M$ can be part of an equivalence. For a functor from $\Ch{\Ab}$ to $\sAb$ we cannot simply take the same collection of abelian groups. This is due to the fact that the degeneracy maps should be injective. This means that for a simplicial abelian group $A$, if we know $A_n$ is non-trivial, then all $A_m$ for $m > n$ are also non-trivial.
But for chain complexes it \emph{is} possible to have trivial abelian groups $C_m$, while there is a $n < m$ with $C_n$ non-trivial. Take for example the chain complex
$$ C =\ldots\to0\to0\to\Z. $$
@ -34,10 +43,10 @@ To repair this defect we should be more careful. Given a simplicial abelian grou
Given a simplicial abelian group $A$, we define abelian groups $N(A)_n$ as
Now define group homomorphisms $\del: N(A)_n \to N(A)_{n-1}$ as
Now define group homomorphisms $\del: N(A)_n \to N(A)_{n-1}$ as
$$\del= d_0|_{N(A)_n}. $$
\begin{lemma}
The function $\del$ is well-defined. Furthermore $\del\circ\del=0$.
@ -52,10 +61,10 @@ The chain complex $N(A)$ is called the \emph{normalized chain complex} of $A$.
\end{lemma}
\begin{proof}
Given a map $f: A \to B$ of simplicial abelian groups, we consider the restrictions
$$ f_n |_{N(A)_n}: N(A)_n \to B_n. $$
$$ f_n |_{N(A)_n}: N(A)_n \to B_n. $$
Because $f_n$ commutes with the face maps we get
$$ d_i(f_n(x))= f_{n-1}(d_i(x))=0, $$
for $i>0$ and $x \in N(A)_n$. So the restriction also restricts the codomain, i.e. $f_n |_{N(A)_n}: N(A)_n \to N(B)_n$ is well-defined. Furthermore it commutes with the boundary operator, since $f$ itself commutes with all face maps. This gives functoriality $N(f): N(A)\to N(B)$.
for $i>0$ and $x \in N(A)_n$. So the restriction also restricts the codomain, i.e. $f_n |_{N(A)_n}: N(A)_n \to N(B)_n$ is well-defined. Furthermore it commutes with the boundary operator, since $f$ itself commutes with all face maps. This gives functoriality $N(f): N(A)\to N(B)$.
@ -125,9 +134,9 @@ We can extend the above lemmas to a more general statement.
\begin{lemma}
\label{le:decomp3}
For all $x \in X_n$ we can write $x$ as:
For all $x \in X_n$ we can write $x$ as
$$ x =\sum_\beta\beta^\ast(x_\beta), $$
for certain $x_\beta\in N(X)_p$ and$\beta : [n]\epi[p]$.
for certain $x_\beta\in N(X)_p$, where $\beta$ ranges over all surjective functions$\beta : [n]\epi[p]$.
\end{lemma}
\begin{proof}
We will proof this using induction on $n$. For $n=0$ the statement is clear because $N(X)_0= X_0$.
@ -136,7 +145,7 @@ We can extend the above lemmas to a more general statement.
\end{proof}
\begin{lemma}
\label{le:decomp4}
Let $\beta: [n]\epi[m]$ and $\gamma : [n]\epi[m']$ be two maps such that $\beta\neq\gamma$. Then we have $\beta^\ast(N(X))_m \cap\gamma^\ast(N(X))_{m'}=0$.
Let $\beta: [n]\epi[m]$ and $\gamma : [n]\epi[m']$ be two maps such that $\beta\neq\gamma$. Then we have $\beta^\ast(N(X))_m \cap\gamma^\ast(N(X))_{m'}=0$.
\end{lemma}
\begin{proof}
Note that $N(X)_i$ only contains non-degenerate $i$-simplices (and $0$). For $x \in\beta^\ast(N(X))_p \cap\gamma^\ast(N(X))_q$ we have $x =\beta^\ast y =\gamma^\ast y'$, where $y$ and $y'$ are non-degenerate. By Lemma~\ref{le:non-degenerate} we know that every $n$-simplex is \emph{uniquely} determined by a non-degenerate simplex and a surjective map. For $x \neq0$ this gives a contradiction.
@ -157,8 +166,8 @@ And by considering $X_n$ as a whole we get:
Using Corollary~\ref{cor:decomp} we can prove a nice categorical fact about $N$, which we will use later on.
\begin{lemma}
The functor $N$ is fully faithful, i.e.:
$$ N: \Hom{\sAb}{A}{B}\iso\Hom{\Ch{\Ab}}{N(A)}{N(B)}\quad A, B \in\sAb. $$
The functor $N$ is fully faithful, i.e.
$$ N: \Hom{\sAb}{A}{B}\iso\Hom{\Ch{\Ab}}{N(A)}{N(B)}\quad A, B \in\sAb. $$
\end{lemma}
\begin{proof}
First we prove that $N$ is injective on maps. Let $f: A \to B$ and assume $N(f)=0$, for $x \in A_n$ we know $x =\sum_\beta\beta^\ast x_\beta$, so
@ -170,7 +179,7 @@ Using Corollary~\ref{cor:decomp} we can prove a nice categorical fact about $N$,
\end{align*}
where we used naturality of $f$ in the second step, and the fact that $x_\beta\in N(A)$ in the last step. We now see that $f(x)=0$ for all $x$, hence $f =0$. So indeed $N$ is injective on maps.
Secondly we have to prove $N$ is surjective on maps. Let $g: N(A)\to N(B)$, define $f: A \to B$ as
Secondly we have to prove $N$ is surjective on maps. Let $g: N(A)\to N(B)$, define $f: A \to B$ as
$$ f(x)=\sum_\beta\beta^\ast g(x_\beta), $$
again we have written $x$ as $x =\sum_\beta\beta^\ast x_\beta$. Clearly $N(f)= g$.
\end{proof}
@ -178,17 +187,17 @@ Using Corollary~\ref{cor:decomp} we can prove a nice categorical fact about $N$,
If we reflect a bit on why the functor $M$ was not a candidate for an equivalence, we see that $N$ does a better job. We see that $N$ leaves out all degenerate simplices, so it is more carefully chosen than $M$, which included everything. In fact, Corollary~\ref{cor:NandD} exactly tells us $M(X)_n = N(X)_n \oplus D_n(X)$.
\subsection{From $\Ch{\Ab}$ to $\sAb$}
For the other way around we actually get a functor for free, via abstract nonsense. Let $F: \sAb\to A$ be any functor, where $A$ is an abelian category. We are after a functor $G: A \to\sAb$, this means that if we are given $C \in A$, we are looking for a functor $G(C): \DELTA^{op}\to\Ab$. Fixing $C$ in the second argument of the $\mathbf{Hom}$-functor gives: $\Hom{A}{-}{C}: A^{op}\to\Ab$, because $A$ is an abelian category. We see that the codomain of this functor already looks good, now if we have some functor from $\DELTA^{op}$ to $A^{op}$, we can precompose, to obtain a functor from $\DELTA^{op}$ to $\Ab$.
For the other way around we actually get a functor for free, via abstract nonsense. Let $F: \sAb\to A$ be any functor, where $A$ is an abelian category. We are after a functor $G: A \to\sAb$, this means that if we are given $C \in A$, we are looking for a functor $G(C): \DELTA^{op}\to\Ab$. Fixing $C$ in the second argument of the $\mathbf{Hom}$-functor gives: $\Hom{A}{-}{C}: A^{op}\to\Ab$, because $A$ is an abelian category. We see that the codomain of this functor already looks good, now if we have some functor from $\DELTA^{op}$ to $A^{op}$, we can precompose, to obtain a functor from $\DELTA^{op}$ to $\Ab$.
Now recall that we have a family of simplicial sets $\Delta[n]$, which are given by the functor $\Delta: \DELTA\to\sSet$. We can apply the free abelian group pointwise, which gives a functor $\Z^{\ast} : \sSet\to\sAb$. And finally we have our functor $F: \sAb\to A$. Composing these gives:
$$ F \Z^{\ast}\Delta: \DELTA\to A. $$
Now recall that we have a family of simplicial sets $\Delta[n]$, which are given by the functor $\Delta: \DELTA\to\sSet$. We can apply the free abelian group pointwise, which gives a functor $\Z^{\ast} : \sSet\to\sAb$. And finally we have our functor $F: \sAb\to A$. Composing these gives:
$$ F \Z^{\ast}\Delta: \DELTA\to A. $$
We can formally regard this functor as a functor from $\DELTA^{op}$ to $A^{op}$. Now combining this with the $\mathbf{Hom}$-functor gives:
Now this is a functor, because it is a composition of functors. Furthermore it is also functorial in the second argument, giving a functor
$$\Hom{A}{F \Z^{\ast}\Delta(-)}{-}: A \to\sAb$$
$$\Hom{A}{F \Z^{\ast}\Delta(-)}{-}: A \to\sAb$$
where we are supposed to fill in the second argument first, leaving us with a simplicial abelian group.
Now we know that $\Ch{\Ab}$ is an abelian group and we have actually two functors $C, N : \sAb\to\Ch{\Ab}$, so we now have functors from $\Ch{\Ab}\to\sAb$. Of course we will be interested in the one using $N$. So we define the functor:
Now we know that $\Ch{\Ab}$ is an abelian group and we have actually two functors $M, N : \sAb\to\Ch{\Ab}$, so we now have functors from $\Ch{\Ab}\to\sAb$. Of course we will be interested in the one using $N$. So we define the functor:
\foreach\i/\j in {2/2, 3/1, 4/0}\draw[->] (m-1-\i) -- node {$f_\j$} (m-2-\i);
\end{tikzpicture}
\Bigg\}\iso C_0, $$
because for $f_1, f_2, \ldots$ there is now choice at all, and for $f_0: \Z\to C_0$ we only have to choose an image for $1\in\Z$. In the next dimension we see:
because for $f_1, f_2, \ldots$ there is now choice at all, and for $f_0: \Z\to C_0$ we only have to choose an image for $1\in\Z$. In the next dimension we see:
We have already seen homology in chain complexes. We can of course now translate this notion to simplicial abelian groups, by assigning a simplicial abelian group $X$ to $H_n(N(X))$. But there is a more general notion of homotopy for simplicial sets, which is also similar to the notion of homotopy in topology. We will define the notion of homotopy groups for simplicial sets.
When dealing with homotopy groups in a topological space $X$ we always need a base-point $\ast\in X$. This is also the case for simplicial sets. We will notate the chosen base-point of a simplicial set $X$ with $\ast\in X_0$. Note that it is a $0$-simplex, but in fact the base-point is present in all sets $X_n$, because we can consider its degenerate simplices $s_0(\ldots(s_0(\ast))\ldots)\in X_n$, we will also denote these elements as $\ast$. Of course in our situation we are concerned with simplicial abelian groups, where there is an obvious choice for the base-point given by the neutral element $0$.
When dealing with homotopy groups in a topological space $X$ we always need a base-point $\ast\in X$. This is also the case for simplicial sets. We will notate the chosen base-point of a simplicial set $X$ with $\ast\in X_0$. More formally, a \emph{pointed simplicial set}$(X, \ast)$ is a simplicial set $X$ together with a $0$-simplex $\ast\in X_0$. By the Yoneda lemma this $0$-simplex corresponds to a map $\Delta[0]\to X$, and any simplex in the image will be denoted by $\ast$. Another way of saying this is that we denote the degenerate simplices $s_0(\ldots(s_0(\ast))\ldots)\in X_n$ as $\ast$. Of course in our situation we are concerned with simplicial abelian groups, where there is an obvious choice for the base-point given by the neutral element $0$.
\subsection{Homotopy groups}
\begin{definition}
@ -58,7 +58,7 @@ Before proving this, one should have a look at figure~\ref{fig:simplicial_eqrel}
\end{proof}
\begin{definition}
Given a simplicial abelian group $X$, we define the $n$-th homotopy group as:
Given a simplicial abelian group $X$, we define the $n$-th homotopy group as
$$\pi_n(X)= Z_n(X)/\sim. $$
\end{definition}
@ -83,7 +83,7 @@ Note that this is an abelian group, because $Z_n(X)$ is a subgroup of $X_n$, and
\end{proof}
\begin{corollary}
For a chain complex $C$ we have $H_n(C)\iso\pi_n(K(C))$
For a chain complex $C$ we have $H_n(C)\iso\pi_n(K(C))$.
\end{corollary}
\begin{proof}
By the established equivalence we have for any chain complex $C$:
@ -91,9 +91,13 @@ Note that this is an abelian group, because $Z_n(X)$ is a subgroup of $X_n$, and
\end{proof}
\subsection{Topology}
In Section~\ref{sec:Constructions} we saw that we can construct a functor $G: \cat{C}\to\sSet$ if we are provided a functor the other way around. If we can define a functor $F: \DELTA\to\Top$, then for any space $X$ we have a simplicial set $\Hom{\Top}{F-}{X}: \DELTA^{op}\to\Set$. In Section~\ref{sec:Chain Complexes}, we already defined the \emph{topological $n$-simplex}$\Delta^n$ and face maps $\delta^i : \Delta^n \mono\Delta^{n+1}$. We can similarly define degeneracy maps $s^i: \Delta^n \to\Delta^{n-1}$ as:
The reader is invited to check the cosimplicial identities himself and conclude that we now have a functor $F: \DELTA\to\Top$, and hence we have a functor $S: \Top\to\sSet$ given by:
In Section~\ref{sec:Chain Complexes}, we already defined the topological $n$-simplex $\Delta^n \in\Top$. We will now relate these spaces to the standard $n$-simplices $\Delta[n]\in\sSet$. We will define a functor $\Delta^-: \DELTA\to\Top$ as follows
The definition of $\Delta^-(\delta_i)$ was already defined in Section~\ref{sec:Chain Complexes} as the face maps $\delta^i: \Delta^n \to\Delta^{n+1}$. So in addition we defined degeneracy maps. The reader is invited to check the cosimplicial identities himself and conclude that we now have a functor $\Delta^-: \DELTA\to\Top$. By composing this with the $\mathbf{Hom}$-functor we obtain a functor $S: \Top\to\sSet$ given by
$$\text{Sing}(X)_n =\Hom{\Top}{\Delta^n}{X}. $$
Recall construction of the singular chain complex in Section~\ref{sec:Chain Complexes}:
Joshua Moerman
11 years ago