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In what follows, a homotopy is a congruence $\simeq$ on a given category. Given such a homotopy, objects $X$ and $Y$ of the given category are homotopy equivalent when there exist morphisms $f\!:X\rightarrow Y$ and $g\!:Y\rightarrow X$ with $gf\simeq id_X$ and $fg\simeq id_Y$.

Q1: Given abstract simplicial complexes $\Delta,\Delta'$ and simplicial maps $f,g\!:\Delta\rightarrow\Delta'$, is there some (nontrivial yet meaningful) notion of a homotopy $h$ or homotopicness $\simeq$ between $f$ and $g$?

Of course this should be a mathematical object with 'finite information' (the usual continuous homotopies between continuous maps $f,g\!: |\Delta|\rightarrow|\Delta'|$ in general cannot be specified by finite amount of data). I'm asking for a discrete analogue of a continuous homotopy between continuous maps. A desired property would also be that if $f\simeq g$, then $H_n(f)=H_n(g)$.

Q2: Let $K$ be a field and denote $K[x_1,\ldots,x_n|p_1,\ldots,p_k]:=K[x_1,\ldots,x_n]/(p_1,\ldots,p_k)$ for given polynomials $p_1,\ldots,p_k\in K[x_1,\ldots,x_n]$. Given morphisms of $K$-algebras $f,g\!: K[x_1,\ldots,x_m|p_1,\ldots,p_k]\rightarrow K[y_1,\ldots,y_n|q_1,\ldots,q_l]$, is there some (nontrivial yet meaningful) notion of a homotopy $h$ or homotopicness $\simeq$ between $f$ and $g$? What about the special case when $K=\mathbb{Z}_2,\mathbb{Q},\mathbb{R},\mathbb{C}$?

Q3: If the answer to Q2 is no, then is there any meaningful notion of a homotopy equivalence between $K[x_1,\ldots,x_m|p_1,\ldots,p_k]$ and $K[y_1,\ldots,y_n|q_1,\ldots,q_l]$? Such a homotopy equivalence is required to satisfy that if $\Delta$ and $\Delta'$ are homotopy equivalent simplicial complexes, then their Stanley-Reisner rings $K[\Delta]$ and $K[\Delta']$ are also homotopy equivalent. What about the special case when $K=\mathbb{Z}_2,\mathbb{Q},\mathbb{R},\mathbb{C}$? What about the special case when $p_i,q_j$ are square-free monomials?

Basically I'd like to know how the homotopy equivalence of simplicial complexes $\Delta,\Delta'$ affects the rings $K[\Delta],K[\Delta']$. For example, $(n\text{-simplex})\mathbb{B}^n\!\simeq\!\mathbb{B}^0\simeq \mathbb{I}_n(n\text{-path})$, so what algebraic property do $K[\mathbb{B}^n]=K[x_0,\ldots,x_n]$ and $K[\mathbb{B}^0]=K[x_0]$ and $K[\mathbb{I}_n]=K[x_0,\ldots,x_n|x_ix_j; i\!-\!j\!\geq\!2]$ share, whilst $\mathbb{B}^n\!\not\simeq\!\mathbb{S}^n$, so what algebraic property don't $K[\mathbb{B}^n]=K[x_0,\ldots,x_n]$ and $K[\mathbb{S}^n]=K[x_0,\ldots,x_n|x_0\cdots x_n]$ share?

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Q1: yes, but I don't know a reference for this. Two simplicial maps $h, k : K \to L$ are contiguous if for each simplex $\sigma$ of $K$, there is a simplex $\sigma'$ of $L$ such that both $h(\sigma)$ and $k(\sigma)$ are faces of $\sigma'$. The following hold (I was given these as exercises in an algebraic topology course):

  • Any two simplicial approximations of a continuous map $f : |K| \to |L|$ (where $|K|$ denotes the geometric realization) are contiguous.
  • Any two contiguous maps induce homotopic maps $|K| \to |L|$.
  • If $f, g : |K| \to |L|$ are homotopic, then there is a barycentric subdivision $K^{(N)}$ of $K$ and a sequence of simplicial maps $h_1, ... h_n : K^{(N)} \to L$ such that $h_1$ is a simplicial approximation to $f$, $h_n$ is a simplicial approximation to $g$, and each pair $(h_i, h_{i+1})$ is contiguous.

So the third condition can be taken as a combinatorial definition of homotopy. There is a better definition if we work with simplicial sets instead of simplicial complexes; see simplicial homotopy.

Q2: see $\mathbb{A}^1$-homotopy theory.

Q3: my impression is that simplices enter homotopy theory for a reason unrelated to the reason that simplices enter combinatorial commutative algebra. I could be wrong, though.

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  • $\begingroup$ Q1: Any book references for simplicial homotopy theory? Q2: I've found $\mathbb{A}^1$ Algebraic Topology over a Field, Morel, 2012, and Motivic Homotopy Theory, Dundas, Levine, Østvær, Röndigs, Voevodsky, 2007. Any other references for the beginner? Q3: what is the best reference to study all possible cell complexes and all operations on them? Is there anything better/more recent/more complete than The Topology of CW Complexes, Lundell, Weingram, 1969? $\endgroup$ – Leo Jun 11 '13 at 20:17
  • $\begingroup$ Q4: Given topological spaces $X$ and $Y$ and continuous maps $f,g\!:X\rightarrow Y$, does every chain homotopy between singular chain maps $f_\ast,g_\ast\!: S_\ast(X)\rightarrow S_\ast(Y)$ originate from a continuous homotopy between $f,g$? $\endgroup$ – Leo Jun 11 '13 at 20:18
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For Q2 I've heard the following definition (some years ago, so I might be wrong). The idea is to replace the unit interval $[0,1]$ in the usual definition of homotopy by the coordinate ring of the affine line, i.e. $\mathcal{O}(\mathbb{A}^1) = \mathbb{Z}[x]$. Thus, two ring homomorphisms $f,g : R \to S$ are called homotopic if there is a ring homomorphism $h : R \to S[x]$ such that $h(r)(0)=f(r)$ and $h(r)(1)=g(r)$.

Note that this relation is compatible with composition of ring homomorphisms (in the sense $f \simeq g \Rightarrow fs \simeq gs$ and $tf \simeq tg$), reflexive and symmetric (use $S[x] \to S[x], x \mapsto 1-x$). But it is not transitive, see below. In order to get a good notion of homotopy, one has to take the transitive closure. This is then a congruence relation on the category of rings.

The same definition applies to the category of $k$-algebras for an arbitrary (commutative) ring $k$. More generally, one may call two morphisms of $k$-schemes $f,g : X \to Y$ homotopic if there is a morphism of $k$-schemes $h : X \times \mathbb{A}^1 \to Y$ such that $h(-,0)=f$ and $h(-,1)=g$. Also note that using $C([0,1])$ instead of $\mathcal{O}(\mathbb{A}^1)$, one gets the well-known notion of homotopy between homomorphisms of Banach algebras (see e.g. here).

Transitivity. If $S$ is a commutative ring, the following are equivalent:

  1. Homotopy is a transitive relation on $\hom(R,S)$ for every $R$.
  2. There is a ring homomorphism $H : S[x] \times_{e_1,S,e_0} S[x] \to S[x]$ such that $e_0 H = e_0 p_1$ and $e_1 H = e_1 p_2$ (here, $e_i : S[x] \to S$ are given by $s \mapsto s$ and $x \mapsto i$, therefore $S[x] \times_{e_1,S,e_0} S[x] = \{(p,q) \in S[x] \times S[x] : p(1)=q(0)\}$).
  3. There are two polynomials $p,q \in S[x]$ such that $pq=0$ and $p(0)=-1$, $p(1)=0$ and $q(0)=0$, $q(1)=1$.

In particular, this fails when $S$ is an integral domain.

Proof:

$1. \Rightarrow 2.$ Let $R=S[x] \times_{e_1,S,e_0} S[x]$. Then $e_0 p_1 \simeq e_1 p_1 = e_0 p_2 \simeq e_1 p_2$, hence also $e_0 p_1 \simeq e_1 p_2$, which is exactly 2.

$2. \Rightarrow 1.$ Let $f,g,h \in \hom(R,S)$ with $f \simeq g \simeq h$, say via $H_1,H_2 \in \hom(R,S[x])$ with $e_0 H_1 = f$, $e_1 H_1 = g$ and $e_0 H_2 = g$, $e_1 H_2 = h$. Because of $e_1 H_1 = e_0 H_2$ there is a unique $T \in \hom(R,S[x] \times_{e_1,S,e_0} S[x])$ with $p_1 T = H_1$ and $p_2 T = H_2$. Then $H_3 := H T \in \hom(R,S[x])$ satisfies $e_0 H_3 = f$ and $e_1 H_3 = h$ and therefore witnesses $f \simeq h$.

$2. \Leftrightarrow 3.$ There is an isomorphism of $S$-algebras $S[u,v]/(uv) \to S[x] \times_{e_1,S,e_0} S[x]$ given by $u \mapsto (x-1,0)$ and $v \mapsto (0,x)$. You can check this directly, or derive it from the algebro-geometric observation that the union of the two coordinate axes is a pushout or gluing of two affine lines at two points. Hence. $2.$ is equivalent to the existence of a ring homomorphism $H : S[u,v]/(uv) \to S[x]$ such that $e_0(H(u))=e_0(x-1)=-1$, $e_1(H(u))=e_1(0)=0$ and $e_0(H(v))=e_0(0)=0$, $e_1(H(v))=e_1(x)=1$. Using the universal properties of quotients and polynomial rings, we see that this is equivalent to the existence of a ring homomorphism $S \to S[x]$ (but this exists anyway) and two polynomials $p=H(u), q=H(v)$ with the properties as in $3.$ $\square$

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  • $\begingroup$ Mmm, I like this definition, it's really simple to state. Two homomorphisms of commutative unital $R$-algebras $f,g\!:A\rightarrow B$ are homotopic if there exists a $R$-algebra homomorphism $h\!:A\rightarrow B[x]$ with $\forall a\!\in\!A\!: h(a)(0)\!=\!f(a),h(a)(1)\!=\!g(a)$, right? For $\simeq$ to be symmetric, we need $B$ to be commutative, so that substitution $B[x]\rightarrow B[x],p(x)\mapsto p(q(x))$ is a $R$-algebra homomorphism, yes? But how can I see compativility with $\circ$? $\endgroup$ – Leo Jun 12 '13 at 1:53
  • $\begingroup$ If $f,g\!:A\rightarrow B$ and $f',g'\!:B\rightarrow C$, with $h\!:A\rightarrow B[x]$, $h\!:f\simeq g$ and $h'\!:B\rightarrow C[y]$, $h'\!:f'\simeq g'$, what is $h''\!:A\rightarrow C[y]$ that gives $h''\!: f'f\simeq g'g$? Perhaps $h'_x(h_x(a))$, but I don't really know what this means, since $h'_x(-)$ must receive an element of $B$, not of $B[x]$. Should we apply $h'_x(-)$ to each coefficient of $h_x(a)$? Furthermore, do you have any counterexample where transitivity indeed fails. And do you know any source where this concept has been developed thoroughly and put to use? $\endgroup$ – Leo Jun 12 '13 at 2:28
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    $\begingroup$ 1. We don't need commutativity for symmetry. Note that $B[x] \cong B[x], x \mapsto 1-x$ is an involutive isomorphism of every $R$-algebra $B$. But actually in my answer I only had commutative rings in mind. 2. The following compatibilities with composition are easy to check: If $f \simeq g$, then $fs \simeq gs$ and $tf \simeq tg$ for all $s,t$ (such that the compositions make sense). I don't know right now if $f \simeq g, f' \simeq g' \Rightarrow ff' \simeq gg'$ holds, but this would then follow from transitivity. 3. I will think about transitivity and try to find sources. $\endgroup$ – Martin Brandenburg Jun 12 '13 at 8:36
  • $\begingroup$ Thank you for your input, and discussion of transitivity! $\endgroup$ – Leo Jun 13 '13 at 17:43

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