@ -59,9 +59,32 @@ Note that we will often drop the indices of the boundary morphisms, since it is
Given a chain complex $C$ we define the \emph{$n$-th homology group}$H_n(C)$ for each $n \in\N$ as:
Given a chain complex $C$ we define the \emph{$n$-th homology group}$H_n(C)$ for each $n \in\N$ as:
$$ H_n(C)= Z_n(C)/ B_n(C).$$
$$ H_n(C)= Z_n(C)/ B_n(C).$$
\end{definition}
\end{definition}
\begin{lemma}
The $n$-th homology group gives a functor $H_n : \Ch{\Ab}\to\Ab$ for each $n \in\N$.
\end{lemma}
\begin{proof}
Let $f: C \to D$ be a chain map and $n \in\N$. First note that for $x \in Z_n(X)$ we have $\del^C(x)=0$, so $\del^D(f_n(x))=0$, because the square on the right commutes:
{\centering
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=2em, column sep=2em]{
\cdots& C_{n+1}& C_n & C_{n-1}&\cdots\\
\cdots& D_{n+1}& D_n & D_{n-1}&\cdots\\
};
\foreach\i/\j in {1/2,2/3,3/4,4/5}\path[->] (m-1-\i) edge node[auto] {$\del^C $} (m-1-\j);
\foreach\i/\j in {1/2,2/3,3/4,4/5}\path[->] (m-2-\i) edge node[auto] {$\del^D $} (m-2-\j);
So there is an induced group homomorphism $f^Z_n : Z_n(C)\to Z_n(D)$ (for $n=0$ this is trivial). Similarly there is an induced group homomorphism $f^B_n : B_n(C)\to B_n(D)$ by considering the square on the left. Now define the map $H_n(f) : x \mapsto[f_n(x)]$ for $x \in Z_n(C)$, we now know that $f_n(x)$ is also a cycle, because of $f^Z_n$. Furthermore it is well-defined on classes, because of $f^B_n$. So indeed there is an induced group homomorphism $H_n(f) : H_n(C)\to H_n(D)$.
It remains to check that $H_n$ preserves identities and compositions. By writing out the definition we see $H_n(\id)([x])=[\id(x)]=[x]=\id[x]$, and:
In order to see why we are interested in the construction of homology groups, we will look at an example from algebraic topology. We will see that homology gives a nice invariant for spaces. So we will form a chain complex from a topological space $X$. In order to do so, we first need some more notions.
In order to see why we are interested in the construction of homology groups, we will look at an example from algebraic topology. We will see that homology gives a nice invariant for spaces. So we will form a chain complex from a topological space $X$. In order to do so, we first need some more notions.
\begin{definition}
\begin{definition}
@ -104,11 +127,11 @@ We now have enough tools to define the singular chain complex of a space $X$.
This might seem a bit complicated, but we can pictures this in an intuitive way, as in figure~\ref{fig:singular_chaincomplex3}. And we see that the boundary operators really give the boundary of an $n$-simplex. To see that this indeed is a chain complex we have to proof that the composition of two such operators is the zero map.
This might seem a bit complicated, but we can pictures this in an intuitive way, as in figure~\ref{fig:singular_chaincomplex}. And we see that the boundary operators really give the boundary of an $n$-simplex. To see that this indeed is a chain complex we have to proof that the composition of two such operators is the zero map.
@ -8,16 +8,29 @@ Before defining \emph{simplicial abelian groups}, we will first discuss the more
We define a category $\DELTA$, where the objects are the finite ordinals $[n]=\{0, \dots, n\}$ for $n \in\N$ and maps are monotone increasing functions: $\Hom{\DELTA}{[n]}{[p]}=\{ f : [n]\to[p]\I f(i)\leq f(j)\text{ for all } i < j \}$.
We define a category $\DELTA$, where the objects are the finite ordinals $[n]=\{0, \dots, n\}$ for $n \in\N$ and maps are monotone increasing functions: $\Hom{\DELTA}{[n]}{[p]}=\{ f : [n]\to[p]\I f(i)\leq f(j)\text{ for all } i < j \}$.
\end{definition}
\end{definition}
There are two special kinds of maps in $\DELTA$, the so called \emph{face} and \emph{degeneracy} maps, defined as (resp.):
There are two special kinds of maps in $\DELTA$, the so called \emph{face} and \emph{degeneracy} maps. The $i$-th face maps $\delta_i$ is the unique injective monotone increasing 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):
$$\delta_i: [n-1]\to[n], k \mapsto\begin{cases} k &\text{if } k < i;\\ k+1&\text{if } k \geq i. \end{cases}\hspace{0.5cm}0\leq i \leq n. $$
\todo{sAb: Consider changing $i$ to be a superscript}
The $i$-th degeneracy map $\sigma_i$ is the unique surjective monotone increasing function which hits $i$ twice. More precisely it is defined for all $n \in\N$ as:
\todo{sAb: Introduce face/degen maps with text, ``unique monotone such that...''}
$$\sigma_i: [n+1]\to[n], k \mapsto\begin{cases} k &\text{if } k \leq i;\\ k-1&\text{if } k > i. \end{cases}\hspace{0.5cm}0\leq i \leq n. $$
\begin{align*}
\delta_i: [n] \to [n+1], k &\mapsto\begin{cases} k &\text{if } k < i;\\ k+1 &\text{if } k \geq i. \end{cases}\hspace{0.5cm} 0 \leq i \leq n+1, \text{ and}\\
\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{0.5cm} 0 \leq i \leq n
\end{align*}
for each $n \in\N$. The nice things about these maps is that every map in $\DELTA$ can be decomposed to a composition of these maps. \todo{sAb: Epi-mono factorization of $\DELTA$} So in a sense, these are all the maps we need to consider. We can now picture the category $\DELTA$ as in figure~\ref{fig:delta_cat}.
The nice things about these maps is that every map in $\DELTA$ can be decomposed to a composition of these maps. So in a sense, these are all the maps we need to consider.
\begin{lemma}
\label{le:epimono}
Let $\eta : [m]\to[n]$ be a map in $\DELTA$. Then $\eta$ can be uniquely decomposed as:
such that $0\leq j_b < \cdots < j_1 < m$ and $0\leq i_1 < \cdots < i_a \leq n$.
\end{lemma}
\begin{proof}
We start with the existence. Consider the set $S =\{ k \in[m-1]\I\eta(k)=\eta(k+1)\}$. These are precisely the elements which are hit twice, now let $S =\{ j_1, \ldots, j_{|S|}\}$ with $0\leq j_{|S|} < \cdots < j_1 < m$. This gives rise to a surjection $\sigma=\sigma_{j_b}\cdots\sigma_{j_1}: [m]\epi[m-|S|]$.
Similarly consider $T =\{ k \in[m - |S|]\I k \not\in\eta[m]\}$. These are precisely the elements which are omitted, now let $T =\{ i_1, \ldots, i_{|T|}\}$ with $0\leq i_1 < \cdots < i_{|T|}\leq n$. This gives an injection $\delta=\delta_{i_a}\cdots\delta_{i_1} : [m - |S|]\mono[n]$. Now we see that $\eta=\delta\sigma$.
Now for uniqueness, \todo{sAb: uniqueuness epi-mono}
\end{proof}
We can now picture the category $\DELTA$ as in figure~\ref{fig:delta_cat}. Note that the face and degeneracy maps are not unrelated. We will make the exact relations precise later.
\begin{figure}[h!]
\begin{figure}[h!]
\includegraphics{delta_cat}
\includegraphics{delta_cat}
@ -25,7 +38,7 @@ for each $n \in \N$. The nice things about these maps is that every map in $\DEL
\label{fig:delta_cat}
\label{fig:delta_cat}
\end{figure}
\end{figure}
Althoug this is a very abstract definition, a more geometric intuition can be given. In $\DELTA$ we can regard $[n]$ as an abstract version of the $n$-simplex $\Delta^n$. The face maps $\delta_i$ are then exactly maps which point out how we can embed $\Delta^n$ in $\Delta^{n+1}$. This is shown in figure~\ref{fig:delta_cat_geom}. This picutre shows the images of the face maps, for example the image of $\delta_3$ from $[2]$ to $[3]$ is the set $\{0,1,2\}$, which is the bottom face of the tetrahedron. The degeneracy maps are harder to visualize, one can think of them as collapsing maps, where two points are identified with each other. \todo{sAb: how to draw $\sigma_i$?}
Although this is a very abstract definition, a more geometric intuition can be given. In $\DELTA$ we can regard $[n]$ as an abstract version of the $n$-simplex $\Delta^n$. The face maps $\delta_i$ are then exactly maps which point out how we can embed $[n]$ in $[n+1]$. This is shown in figure~\ref{fig:delta_cat_geom}. This picutre shows the images of the face maps, for example the image of $\delta_3$ from $[2]$ to $[3]$ is the set $\{0,1,2\}$, which is the bottom face of the tetrahedron. The degeneracy maps are harder to visualize, one can think of them as ``collapsing'' maps, where two points are identified with each other. For example, this collapses a triangle into a line.
\begin{figure}
\begin{figure}
\includegraphics{delta_cat_geom}
\includegraphics{delta_cat_geom}
@ -41,7 +54,7 @@ This category $\DELTA$ will act as a protoype for these kind of geometric struct
(Or equivalently a contravariant functor $X: \DELTA\to\Set.$)
(Or equivalently a contravariant functor $X: \DELTA\to\Set.$)
\end{definition}
\end{definition}
So the category of all simplicial sets, $\sSet$, is the functor category $\Set^{\DELTA^{op}}$, where morphisms are natural transformations. Because the face and degeneracy maps give all the maps in $\DELTA$ it is sufficient to define images of $\delta_i$ and $\sigma_i$ in order to define a functor $X: \DELTA^{op}\to\Set$. And hence we can picture a simplicial set as done in figure~\ref{fig:simplicial_set}. Comparing this to figure~\ref{fig:delta_cat} we see that the arrows are reversed, because $X$ is a contravariant functor.
So the category of all simplicial sets, $\sSet$, is the functor category $\Set^{\DELTA^{op}}$, where morphisms are natural transformations. Because the face and degeneracy maps give all the maps in $\DELTA$ it is sufficient to define images of $\delta_i$ and $\sigma_i$ in order to define a functor $X: \DELTA^{op}\to\Set$, keep in mind that these should satisfy some relations which we will discuss next. Hence we can picture a simplicial set as done in figure~\ref{fig:simplicial_set}. Comparing this to figure~\ref{fig:delta_cat} we see that the arrows are reversed, because $X$ is a contravariant functor.
\begin{figure}
\begin{figure}
\includegraphics{simplicial_set}
\includegraphics{simplicial_set}
@ -84,24 +97,34 @@ It is already indicated that a functor from $\DELTA^{op}$ to $\Set$ is determine
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.
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.
\todo{sAb: Note that $s_i$ is a monomorphism because of (3)}
Note that because of the third equation, the degeracy 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.
\todo{sAb: Degenerate simpl. lemma about non-deg simpl.}
\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 $i$. An element is called \emph{non-degenerate} if this is not the case.
\end{definition}
\begin{lemma}
We can write any $x \in X_n$ uniquely as $x =\beta^\ast y$ for some surjective map $\beta : [n]\epi[m]$ and $y \in X_m$ non-degenerate.
\end{lemma}
\begin{proof}
We will proof the existence with inducion 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 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 $y = z$.
\todo{sAb: $\gamma=\beta$}
\end{proof}
\subsection{The standard $n$-simplex}
\subsection{The standard $n$-simplex}
There are very important simplicial sets:
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:
Note that this is also the definition of the Yoneda embedding $\Delta[n]= y[n]$. In a moment we will see why the Yoneda lemma is useful to us. But we will explicitly describe two of such standaard $n$-simplices.
Note that indeed $\Hom{\DELTA}{[k]}{[n]}\in\Set$, because the collection of morphisms in a category is per definition a set. We do not need to specify the face or degeneracy maps, as we already know that $\mathbf{Hom}$ is a functor (in both arguments). Still it is useful to write out some cases.
\todo{sAb: In the examples note the non-deg simpl.}
\begin{example}
\begin{example}
We will compute how $\Delta[0]$ look like. Note that $[0]$ is an one-element set, so for any set $X$, there is only one function $\ast : X \to[0]$. Hence $\Delta[0]_n =\{\ast\}$ for all $n$. The face and degeneracy maps are now functions from $\{\ast\}$ to $\{\ast\}$. Again there is only one, namely $\id : \{\ast\}\to\{\ast\}$. This gives:
We will compute how $\Delta[0]$ look like. Note that $[0]$ is an one-element set, so for any set $X$, there is only one function $\ast : X \to[0]$. Hence $\Delta[0]_n =\{\ast\}$ for all $n$. The face and degeneracy maps are now functions from $\{\ast\}$ to $\{\ast\}$. Again there is only one, namely $\id : \{\ast\}\to\{\ast\}$. This gives:
In this simplicial set there are three non-degenerate simplices. There is $\id\in\Delta[1]_1$, which clearly is non-degenerate, and the two $0$-simplices $\delta_0$ and $\delta_1$. One can think of this simplicial set as a line (the non-degenerate $1$-simplex) with its endpoints (the two $0$-simplices).
\end{example}
\end{example}
\subsection{Other simplicial objects}
\subsection{Other simplicial objects}
@ -159,7 +183,7 @@ As we are interested in simplicial abelian groups, it would be nice to make thes
\end{figure}
\end{figure}
\begin{example}
\begin{example}
We can apply this to the standard $n$-simplex $\Delta[1]$. This gives $\Delta[1]_0\iso\Z^2$, since $\Delta[1]_0$ had two elements, and $\Delta[1]_1\iso\Z^3$, where the isomorphisms are taken such that:
We can apply this to the standard $n$-simplex $\Delta[1]$. This gives $\Delta[1]_0\iso\Z^2$, since $\Delta[1]_0$ has two elements, and $\Z^\ast[\Delta[1]]_1\iso\Z^3$, where the isomorphisms are taken such that:
\begin{align*}
\begin{align*}
\delta_0 &\mapstot{\iso} (1, 0) \\
\delta_0 &\mapstot{\iso} (1, 0) \\
\delta_1 &\mapstot{\iso} (0, 1) \\
\delta_1 &\mapstot{\iso} (0, 1) \\
@ -167,9 +191,17 @@ As we are interested in simplicial abelian groups, it would be nice to make thes
\id&\mapstot{\iso} (0, 1, 0) \\
\id&\mapstot{\iso} (0, 1, 0) \\
\sigma_0\delta_1 &\mapstot{\iso} (0, 0, 1)
\sigma_0\delta_1 &\mapstot{\iso} (0, 0, 1)
\end{align*}
\end{align*}
The face maps from $\Delta[1]_1$ to $\Delta[1]_0$ under these isomorphisms are then given by:
The face maps from $\Z^\ast[\Delta[1]]_1$ to $\Z^\ast[\Delta[1]]_0$ under these isomorphisms are then given by:
\begin{align*}
\begin{align*}
d_0(x, y, z) &= (x+y, z) \\
d_0(x, y, z) &= (x+y, z) \\
d_1(x, y, z) &= (x, y+z)
d_1(x, y, z) &= (x, y+z)
\end{align*}
\end{align*}
\end{example}
\end{example}
\subsection{The Yoneda lemma}
Recall that the Yoneda lemma stated: $\mathbf{Nat}(y(C), F)\iso F(C)$, where $F:\cat{C}^{op}\to\Set$ is a functor and $C$ an object. In our case we consider functors $X: \DELTA^{op}\to\Set$ and objects $[n]$. So this gives us the natural bijection:
$$ X_n \iso\Hom{\sSet}{\Delta[n], X}. $$
So we can regard $n$-simplices in $X$ as maps from $\Delta[n]$ to $X$. This also extends to the abelian case, where we get an natural isomorphism (of abelian groups):
Joshua Moerman
12 years ago
