(v.1.1)
Let $h(CW)$ be the homotopy category of CW-complexes. Show that this category does not have all pushouts. More concretely, show that the diagram \[ *\xleftarrow{} S^1\xrightarrow{\times 2} S^1 \] does not admit a pushout (great example).
proof:
In point set $CW$, the pushout is just $*$, as $\times 2$ is surjective. However, the diagram is also equivalent to \[ D^1\xleftarrow{} S^1\xrightarrow{\times 2} S^1 \] With pushout in point set $\mathbb{R}P^2$.
Show that every map in $\Delta$ can be uniquely factored as a composition of $s_i $’s followed by a composition of $d_j $’s. Therefore, a simplicial set is equivalently described by a sequence of sets $X_n$ equipped with face and degeneracy maps satisfying the simplicial identities.
proof:
Let $f:[n]\to [m]$ be a map of ordered set. We have that the map factors through $im (f)$, a composition of surjective order preserving map, and injective order preserving map. For the surjective side, we may obtain the map through a list of composition of $s_i$’s that combine the adjacent two elements. For the injective side, we may introduce more element in between two elements.
Give examples of simplicial sets where the relation of Definition 1.1.9 (i.e. $x \sim y \in X_0$ if there is $H\in X_1$ such that $d_0(H)=x$ and $d_1(H)=y$), leading to $\pi_0^\Delta (X)$, is not symmetric and not transitive.
proof:
Let $X=\Lambda^2_1$. $0\sim 1$ but there is no $1$-simplice such that $1\sim 0$, so not symmetric. $0\sim 1$ and $1\sim 2$, but there is no $1$ -simplice such that $0\sim 2$, so not transitive.
Show that every simplex $x \in X_m$ is of the form $\alpha_*(y)$ for a surjection $\alpha : [m] \to [n]$ and a non-degenerate $n$-simplex $y$, and show that the pair $(\alpha, y)$ is uniquely determined by $x$ (another classic result).
proof:
First of all, if there is no such $y$ for $n< m$, then $x$ is nondegenerate. So it remains to show the uniqueness.
Suppose we have another $(\beta, y’)$ for some $\beta:[m]\to [n’]$ surjection. Suppose that $n’\geq n$, then there is a surjection $f: [n’]\to[n]$ such that $f\circ \beta =\alpha$. There is another map $g:[n’]\to [m]$ such that $\beta\circ g= id$. Hence \[ X(f)(y)=X(g)X(\alpha)(y)=X(g)(x)=X(g)X(\beta)(y’)=y’ \] Hence $y’$ is either nondegenerate, or $n’=n$, but this implies $f=id$.
Let $F : I \to C$ be a functor. Show that a colimit of $F$ can equivalently be described as an initial cocone over $F$, and that a limit of $F$ can be equivalently described as a terminal cone over $F$.
proof:
Showing the colimit case suffices. Suppose $F^{\triangleright}$ be the initial cocone, then any other cocone $G$, such that $G|_{I}=F$, there exists an unique morphism $F^{\triangleright}\to G$, which corresponds to an unique map that determines the universal property of colimit.
Calculate the limit and colimit of a simplicial set $X : \Delta^{op}\to Set$.
proof:
The computation using equalizer and coequalizer come in handy. We have that the limit will be the elements that are equal in $X_0$ under all morphisms, hence $X_0$. The colimits is the equivalent classes of elements that share vertices, hence the connected components.
Show that the datum of an adjunction in the sense of Definition 1.1.17(natural isomorphism $\alpha: \Hom(F(-), -)\to \Hom(-,G(-))$) is equivalent to the datum of a pair of functors $(F,G)$ together with natural transformations $\epsilon : F \circ G \to id$ and $\eta : id \to G\circ F $ satisfying the triangle identities, i.e., the obvious composites \[ F\xrightarrow{F\eta } FGF\xrightarrow{\epsilon F} F \]
\[ G\xrightarrow{\eta G} GFG\xrightarrow{G\epsilon } G \] are the identities
proof:
Suppose we have the natural transformations, then \[ \Hom(F(-),-)\xrightarrow{G} \Hom(GF(-), G(-))\xrightarrow{\eta^{*}}\Hom(-,G(-)) \] and \[ \Hom(-,G(-))\xrightarrow{F} \Hom(F(-), FG(-))\xrightarrow{\epsilon_{*}}\Hom(F(-),-) \] are inverses to each other.
$\alpha$ determines than natural isomorphisms, \[ \alpha_{FG}: \Hom(FG(-), -)\to \Hom(G(-),G(-)) \] and \[ \alpha_{GF}: \Hom(F(-), F(-))\to \Hom(-,GF(-)) \] Therefore, we have the identity of $F$ and $G$ resp. determines natural transformation $\epsilon : F \circ G \to id$ and $\eta : id \to G\circ F $.
\[ \Hom(F(-), F(-))\xrightarrow{F\alpha_{GF}} \Hom(F(-),FGF(-))\xrightarrow{\epsilon F_{*}} \Hom(F(-), F(-)) \] Takes $id:F(X)\to F(X)$ to $\epsilon F\circ F\eta=\epsilon F\circ \alpha_{GF}(id_{Fx})=\alpha_{GF}^{-1}(\alpha_{GF}(id_{Fx}))=id_{Fx}$
Show that $F:C\to D$ admits a right adjoint iff there is $G: D\to C$ such that for each $y\in D$, we have a morphism $\epsilon_y: FGy\to y$ such that for any $x\in C$, there is a bijection \[ \Hom(x, Gy)\xrightarrow{F}\Hom(Fx, FGy)\xrightarrow{\epsilon_y \circ}\Hom(Fx, y) \] proof:
If $G$ is a right adjoint, then Exercise 9 suffice. So we prove the other way.
We see that the bijection is the same as for any $x\in C$, and map $f:Fx\to y$, then it factors through $FGy$ uniquely via a map $x\to Gy$.
Suppose, $y’\to y$ is a morphism, then $FGy’\to y’\to y$ factors through $FGy$ uniquely with $Gy’ \to Gy$, since it is determined by a bijection, it comes from $y’\to y$.
Hence $\epsilon$ is actually a natural transformation. Then it is just definition of adjunction.
Prove that if a simplicial set $X$ has at most $n$-dimensional non-degenerate simplices, and $Y$ has at most $m$-dimensional non-degenerate simplices, then their product $X \times Y$ has at most $(n + m)$-dimensional non-degenerate simplices.
proof:
For $(x,y)\in (X\times Y)_{n+m+1}$, we have $x\in X_{n+m+1}$ and $y\in Y_{n+m+1}$. Let $(\alpha, x’)$ and $(\beta, y’)$ be the uniqe nondegenerate simplice for respectively $x$ and $y$. Since $|x|\leq n$ and $|y|\leq m$, then there exists $k\leq n+m$, a surjection $[n+m+1]\to [k]$, and surjection $\alpha’:[k]\to [|x|]$ and $\beta’$ such that $\alpha=\alpha’\circ f$ and $\beta=\beta’\circ f$. Hence $(x,y)$ is degenerate.
Show that for every simplicial set $X$, there is a canonical bijection $\pi_0^\Delta(X)\cong \pi_0(|X|)$
proof:
Define $f:\pi_0^\Delta(X)\cong \pi_0(|X|)$ from $x\in X_0$ to $(x,*)\in |X|$.
Since if $x\sim y$ in $X_0$, then there is a $\Delta[1]\to X$ connecting $x$ to $y$, so realization of $x$ and $y$ lies on the same connected component, so this is well defined.
We have that for every connected component of $|X|$, there is at least one $x\in X_0$, so $f$ is surjective.
If we have $(x,*)$ and $(y,*)$ lying on the same connected component. Then there exists a zigzag of $\Delta[1]$ connecting them in $X$, so $x\sim y$.
Show there is pushout diagram \begin{array}{ccc} \coprod_{J_n}\partial \Delta[n]& \longrightarrow & \mathrm{sk}_{n-1}(X) \end{array} \begin{array}{ccc} \downarrow && \downarrow \end{array} \begin{array}{ccc} \coprod_{J_n}\partial \Delta[n] & \longrightarrow & \mathrm{sk}_{n}(X) \end{array} Where $J_n$ are the nondegenerate $n$-simplices. Moreover, $\mathrm{colim}_n\mathrm{sk}_n(X)\cong X$.
proof:
The diagram above is defined through the inclusion $\coprod_{J_n}\partial \Delta[n] \to \mathrm{sk}_{n}(X)$ and canonical $\mathrm{sk}_{n-1}(X)\to \mathrm{sk}_{n}(X)$. It is obvious that this is a pushout diagram.
There is a sequence $\mathrm{sk}_n(X)\to \mathrm{sk}_{n+1}(X)\to…\to X$, where $\mathrm{sk}_n(X)\to \mathrm{sk}_{n+1}(X)$ is injection. Suppose we have a $\mathrm{sk}_n(X)\to \mathrm{sk}_{n+1}(X)\to … \xrightarrow{h} Y$. For every, $n$-simplice $x\in X_n$, there is a large enough $m\geq n$ such that $x\in\mathrm{sk}_m(X)_n$. Hence, $h$ determines a simplicial map $f: X\to Y$. Suppose, there are two $f, g:X\to Y$, that factors $h$. Given $x\in X_n$, $x\in \mathrm{sk}_m(X)_n$ for some $m$, and $f|_{\mathrm{sk}_m(X)}=g|_{\mathrm{sk}_m(X)}$, so $f(x)=g(x)$. Hence $f=g$.
Show that the following simplicial sets are not nerves of categories:
We will use the fact that a simplicial set is a nerve if it has uniques inner horn fillers.
$\partial \Delta[n]$ for $n\geq 2$.
proof:
Obviously, $Lambda^n_j\to \partial \Delta[n]$ does not have filler for any $j$, so not a nerve
$\Lambda^n_i$ for $n= 2$, $j=1$ and $n>2$, $0\leq j\leq n$ proof:
For $n=2$, $\Lambda^2_1=I^2$, we leave this for next one. For $n>3$, there is embedding $\partial \Delta[n-1]\to \Lambda^n_j$ that does not factor through $\Delta[n-1]$.
$I^n$ for $n\geq 2$
proof:
$I^2=\Lambda^2_1$ has no filler, and $I^2$ embeds into $I^n$ that does not factor through $\Delta[2]$.
Suppose that $X$ is a Kan complex. Show that for all $n \geq 0$, the simplicial set $\mathrm{cosk}_n(X)$ is again a Kan complex. Prove that the canonical map $X \to \mathrm{cosk}_n(X)$ induces a bijection \[ \pi_k(X)\to \pi_k(\mathrm{cosk}_n(X)) \] For $k< n$, and $\pi_k(\mathrm{cosk}_n(X))=0$ for $k\geq n$
proof:
The function \[ \Hom(\Delta[k], \mathrm{cosk}_n(X))\to \Hom(\Lambda^k_i, \mathrm{cosk}_n(X)) \]is isomorphic to \[ \Hom(\mathrm{sk}_n\Delta[k], X)\to \Hom(\mathrm{sk}_n\Lambda^k_i, X) \] So we need to show that this is a surjection.
For $k<n-1$, $\mathrm{sk}_n\Lambda^k_i\to \mathrm{sk}_n\Delta[k]$ is isomorphism to the horn inclusion $\Lambda^k_i \to \Delta[k]$.
For $k>n-1$, $\mathrm{sk}_n\Lambda^k_i\to \mathrm{sk}_n\Delta[k]$, is an isomorphism.
The only issue is with $k=n-1$, we have the inclusion is actually $\Lambda^k_i\to \partial \Delta[k]$. The fact that $K$ is a Kan complex implies for any $\Lambda^k_i$, we may determine the $\partial \Delta[k]$ filling through the barycentric subdivision.
Hence $\mathrm{cosk}_n(X)$ is Kan.
Let $S^k$ be $k$-dimensional simplicial sphere $\Delta[k]/\partial \Delta[k]$. We have that since $\mathrm{sk}_n$ is a left adjoint, $\mathrm{sk}_n S^k=(\mathrm{sk}_n \Delta[k])/(\mathrm{sk}_n\partial \Delta[k])$. If $k<n$, then $\mathrm{sk}_n S^k\cong S^k$, and $\mathrm{sk}_n (S^k\times I)\cong S^k\times I$ (this is because all nondegenerate simplices here are dimension at most $k+1\leq n$). Hence, the isomorphism of the homotopy group for $k<n$ is shown.
For $k= n$, $\mathrm{sk}_n S^k=S^n$ and $\mathrm{sk}_n (S^k\times I)\cong S^n\cup I \cup S^n$, so $\pi_n(\mathrm{cosk}_n X)=0$.
For $k>n$, $\mathrm{sk}_n S^k=*$ and $\mathrm{sk}_n (S^k\times I)\cong *$, so $\pi_k(\mathrm{cosk}_n X)=0$.
Show that a natural transformation between two functors $f, g : C \to D$ induces a homotopy between $N(f ), N(g) : N(C) \to N(D)$. Use this result to show that conjugation with an element determines a self map of $BG$ which is homotopic to the identity. What does conjugation induce on $\pi_1 (BG)$? Why does this not show that every group is abelian? proof:
A natural transformation $\alpha: f\implies g$ is equivalent to a functor $H: C\times I\to D$, such that $H|_{C\times 0}=f$ and $H|_{C\times 1}=g$. The nerve is then $N(H): N(C)\times I\to N(D)$, which is a homotopy between $N(f)$ to $N(g)$.
For conjugation by $g\in G$, we have $c_g$ takes $*\xrightarrow{h} *$ to $*\xrightarrow{ghg^{-1}} *$. There is an natural transformation, $\alpha: id\implies c_g$, \begin{array}{ccc} *& \longrightarrow{g} & * \end{array} \begin{array}{ccc} \downarrow{h} && \downarrow{ghg^{-1}{} \end{array} \begin{array}{ccc} * & \longrightarrow{g} & * \end{array} So we have $c_g$ is homotopic to $id$.
The action of conjugation induces conjugation on $\pi_1$. Why does this not imply every group is abelian? The homotopy we constructed is not a based homotopy.
Show that the nerve of a category $C$ is 2-coskeletal, i.e., that the canonical map $N(C) \to \mathrm{cosk}_2(N(C))$ is an isomorphism of simplicial sets.
proof:
The canonical map on $n$-simplices $N(C)_n \to \mathrm{cosk}_2(N(C))_n$ is equivalent to the map \[ \Hom(\Delta[n], N(C))\to \Hom(\Delta[n], \mathrm{cosk}_2(N(C)))\cong \Hom(\mathrm{sk}_2\Delta[n], N(C)) \]
$\mathrm{sk}_2\Delta[n]$ contains the $2$-faces, which determines composable edges. Since $\Hom(\Delta[n], N(C))$ is isomorphic to the length $n$-composable edges, which are uniquely determined by the sequences of $2$-composable edges, this is an isomorphism.
Let $G$ be a group and let $BG$ be the category with one object and $G$ as endomorphisms of that object. Show that $N(BG)$ has only one non-trivial homotopy group, namely $\pi_1(N(BG))$, and that this group is canonically isomorphic to $G$.
proof:
First of all, by Exercise 19, we have $N(BG)\cong \mathrm{cosk}_2N(BG)$. Then by Exercise 17, we have $\pi_k(N(BG))=0$ for $k\geq 2$. $N(BG)$ is connected so $\pi_0(N(BG))=0$.
For $f\in\Hom(S^1, N(BG))$, $f()=$ and $f(e)=g\in G$ for $e$ the unique nondegenerate $1$-simplice, and $f$ is uniquely determined by $g$. Suppose we have a based homotopy $H: \Delta[1]^2/\partial \Delta[1]\times \Delta[1]\to N(BG)$, the only three nondegenerate $1$-simplices, point in the same direction, living along adjacent $2$-faces. Hence they must be equal. In order word, $\pi_1(N(BG))=\Hom(S^1, N(BG))=G$
Consider the map $[0] \to [n]$ with image {$0$}. Show that this determines a map $0: [0] \to\partial \Delta[n]$. Calculate the simplicial homotopy sets $\pi_i(\partial \Delta[n])$ for $i \geq 1$ and $n\geq 2$. Deduce that $\partial \Delta[n]$ is not a Kan complex.
proof:
A map $S^n\to \pi_i(\partial \Delta[n])$ based at $0$, has to factor through $[0]$ because the only $n$-simplices with $\partial \Delta[n]=0$ is the degenerate one. Therefore $\pi_i(\partial \Delta[n])=0$. If $\partial \Delta[n])$ is Kan then this implies $\partial \Delta[n])$ is contractible, so $\partial \Delta[n])$ is not Kan.
Determine the homotopy category of the following simplicial sets:
We use Lemma 1.2.7 that says if $X\to Y$ induces isomorphism on $\mathrm{sk}_2X\to \mathrm{sk}_2Y$, then $hX\to hY$ is an isomorphism.
$\partial \Delta[n]$ proof:
For $n>2$, $\mathrm{sk}_2 \partial \Delta[n]=\mathrm{sk}_2\Delta[n$. Therefore, $h\partial \Delta[n]\cong [n]$.
For $n=2$, $\mathrm{sk}_2 \partial \Delta[2]=\partial \Delta[2]$. There are three objects, three arrows, two of which are composable, and no $2$-simplices. Therefore, $h\partial \Delta[2]$ is the category of three objects, $0$, $1$, and $2$, with arrow $0\to 1$, $0\to 2$, and $1\to 2$, with $0\to 1\to 2\neq 0\to 2$.
For $n=1$, $\partial \Delta[1]$ is the disjoint union of two point, so $h\partial \Delta[1] $ is the disjoint category of two object.
$\Lambda^n_j$ for $n\geq 2$ and $0\leq j \leq n$
proof:
It is the category of $n+1$ objects, a $n$-composable sequence of arrows, and an arrow $0\to n$ that is different from $0\to 1\to …\to \hat {j}\to …\to n$.
$I^n$
proof:
$hI^n\cong [n]$.
Let $f : X \to Y$ be a map of simplicial sets. Prove or give a counter example to the following statements:
If $f$ is a monomorphism, then $hX \to hY$ is fully faithful.
counterexamples:
$\partial \Delta[1]\to \Delta[1]$ is mono, but $h\partial \Delta[1]=0\cup 1\to h\Delta[1]=[1]$ is not full.
Let $X:=\partial \Delta[2]/[1\to 2]$ and let $Y:=\Delta[2]/[1\to 2]$, so we identify the edge $1\to 2$. There is a monomorphism $X\to Y$, but $hX$ is the two object category, with two distinct arrows, yet $hY$ has only one arrow, so it is not faithful.
If $f$ is a degree-wise surjection, then $hX \to hY$ is surjective and full, i.e., it induces a surjection on objects and on hom-sets.
proof:
see iii)
If $f$ induces a surjection on $0$- and $1$-simplices, then $hX \to hY$ is surjective and full.
proof:
Since $f$ is surjection. It obviously induces a surjection on the objects. Let $e: x\to y$ in $hY$, it corresponds to an arrow $e’:x’\to y’$ in $hX$, because $e:x\to y$ corresponds to an edge $e:x\to y$ in $Y$.
Let $X$ be a simplicial set and consider the canonical map $X \to N(hX)$.
Show that this map factors through the canonical map $X \to \mathrm{cosk}_2(X)$.
proof:
We have that $X\to \mathrm{cosk}_2(X)$ determines an isomorphism $hX\to h\mathrm{cosk}_2(X)$.
Show that the induced map $ \mathrm{cosk}_2(X) \to N(hX)$ is an isomorphism if $X$ is isomorphic to the nerve of a category.
proof:
If $X=N(C)$, then $N(hX)=X$, and we have that nerve of categories are $2$-coskeletal.
Show that the map $\mathrm{cosk}_2(X) \to N(hX)$ is in general not an isomorphism. Hint: Find an $X$ which is $2$-coskeletal, but not the nerve of a category.
proof:
Let $X$ be the simplicial set of $2$ copies of $\Delta[2]$ glued together along the $0\to 1$ and $1\to 2$. This is obviously not the nerve of category.
For $\Hom(\Delta[n], X)\to \Hom(\mathrm{sk}_2(\Delta[n]), X)$, suppose we have $f:\mathrm{sk}_2(\Delta[n])\to X$, $f$ factors through one copy of $\Delta[2]$. Therefore, there is a filler by $\Delta[n]$, hence $X$ is $2$-coskeletal.
Prove or disprove the following statement: The map $\mathrm{cosk}_2(X) \to N(hX)$ is an isomorphism if and only if $X$ is isomorphic to the nerve of a category.
counterexample:
Let $X$ be the pushout of $\Delta[3]\xleftarrow{}\partial \Delta[3]\to \Delta[3]$. Its $2$-coskeleton is $\Delta[3]$, as $\mathrm{sk}_2 \Delta[n]\to X$ lives within $\partial \Delta[3]$. We have then $\Delta[3]\to N(hX)$ is isomorphism, but $X$ is not a nerve.
Let $(V , \otimes, \mathbb{1})$ be a monoidal category. Then the functor $\Hom_V (\mathbb{1}, −) : V \to Set$ is lax monoidal. Is it monoidal? If not: Can you find a condition on $(V , \otimes, \mathbb{1})$ which ensures that it is?
counterexample:
Let $V=Ab$ with the tensor product, $\mathbb{1}=\mathbb{Z}$. We have then $\Hom_V (\mathbb{1}, −)\cong fgt$, the forgetful functor from abelian groups to sets. We have that $fgt(A\otimes B)\not\cong fgt(A)\times fgt(B)$.
A strong condition on $V$ so the functor is monoidal, if $\otimes$ is the product.