Theory

Simplicial maps, chain maps

Simplicial map

A map $f:\mathcal{K}\to \mathcal{L}$ between simplicial complexes is simplicial if it maps vertices of $\mathcal{K}$ to vertices of $\mathcal{L}$, and it maps simplices of $\mathcal{K}$ linearly to simplices of $\mathcal{L}$.

For example:

• Inclusions of one simplicial complex into another.
• Reduction of a simplicial complex to lower dimension, e.g. $[v_0,v_1]\to [v_0]$.

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Chain map

A chain map $f$ is a map between chain complexes, which consists of a collection $f_n,\dots ,f_0$ of homomorphisms:

$$f_i:C_i(\mathcal{K})\to C_i(\mathcal{L})$$

of abelian groups which commute with the boundary operators:

$$f_{i-1}\circ {\partial }_i^{\mathcal{K}}={\partial }_i^{\mathcal{L}}\circ f_i\text{for all}i.$$

Such a chain map makes the following diagram commute:

• The homomorphisms can be collected under a single map $f$, and the commutativity condition expressed simply:

$$f\circ {\partial }^{\mathcal{K}}={\partial }^{\mathcal{L}}\circ f$$

Theorem: Induced chain map

A simplicial map $f:\mathcal{K}\to \mathcal{L}$ determines a map of chain complexes $f:C(\mathcal{K})\to C(\mathcal{L})$.

The simplicial map $f$ is extended linearly from simplices to chains (from generators to linear combinations).

Induced maps on homology

Theorem: Induced map on homology

A chain map $f:\mathcal{K}\to \mathcal{L}$ determines a map $f_{\ast }$ of homology groups which splits by degrees:

$${f_{\ast }}_i:H_i(\mathcal{K})\to H_i(\mathcal{L})$$

A chain map $f$ sends cycles to cycles and boundaries to boundaries, so it descends to a map of quotient groups.

Proof of Theorem: Induced map on homology

It is sufficient to check that $f$ maps cycles to cycles and boundaries to boundaries:

• By mapping boundaries to boundaries, $f$ is well-defined on homology groups.
• By mapping cycles to cycles, $f$ maps elements of homology to elements of homology.

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Checking $f$ maps cycles to cycles:

If $\tau \in \mathrm{Im}{\partial }^{\mathcal{K}}$, meaning that $\tau$ is a cycle, then $\tau ={\partial }^{\mathcal{K}}(\sigma )$ for some $\sigma$, whence:

$$\tau =(f\circ {\partial }^{\mathcal{K}})(\sigma )\gg \gg \tau =({\partial }^{\mathcal{L}}\circ f)(\sigma ),$$

but this implies $\tau \in \mathrm{Im}{\partial }^{\mathcal{L}}$.

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Checking $f$ maps boundaries to boundaries:

If ${\partial }^{\mathcal{K}}(\tau )=0$, then:

$$\begin{array}{c}{\partial }^{\mathcal{L}}(f(\tau ))\gg \gg ({\partial }^{\mathcal{L}}\circ f)(\tau )\gg \gg (f\circ {\partial }^{\mathcal{K}})(\tau )\\ \\ \gg \gg f({\partial }^{\mathcal{K}}(\tau ))\gg \gg f(0)\gg \gg 0\end{array}$$

but this means the image of $\tau$ is a cycle.

Proposition: Homotopic maps are the same on homology

If two maps of simplicial complexes $f,g:\mathcal{K}\to \mathcal{L}$ are homotopic, then they induce the same maps on homology groups, $f_{\ast }=g_{\ast }$:

In words: continuous deformations do not change the map induced on homology.

Proposition: Homology respects composition of maps

Given a composition of simplicial maps:

$$\mathcal{K}\stackrel{f}{⟶}\mathcal{L}\stackrel{g}{⟶}\mathcal{M},$$

the map induced by the composite agrees with the composite of the induced maps:

$$(g\circ f{)}_{\ast }=g_{\ast }\circ f_{\ast }:H_i(\mathcal{K})\to H_i(\mathcal{M})$$

The property of respecting composition of maps is called functoriality.

Corollary: Homotopy equivalence implies homology isomorphism

Homotopy equivalent simplicial complexes have isomorphic homology groups:

$$\mathcal{K}\simeq \mathcal{L}⟹H_i(\mathcal{K})\cong H_i(\mathcal{L})\text{for all}i$$

Proof of Corollary

Suppose we have a homotopy equivalence:

$$\begin{array}{c}\mathcal{K}\stackrel{f}{⟶}\mathcal{L},\mathcal{L}\stackrel{g}{⟶}\mathcal{K}\\ \\ g\circ f\simeq {\mathrm{Id}}_{\mathcal{K}},f\circ g\simeq {\mathrm{Id}}_{\mathcal{L}}\end{array}$$

Now apply functoriality to the induced maps on homology:

$$\begin{array}{c}{g}_{\ast }\circ f_{\ast }\gg \gg (g\circ f{)}_{\ast }\gg \gg ({\mathrm{Id}}_{\mathcal{K}}{)}_{\ast }={\mathrm{Id}}_{H_i(\mathcal{K})}\\ \\ f_{\ast }\circ g_{\ast }\gg \gg (f\circ g{)}_{\ast }\gg \gg ({\mathrm{Id}}_{\mathcal{L}}{)}_{\ast }={\mathrm{Id}}_{H_i(\mathcal{L})}\end{array}$$

Therefore $g_{\ast }$ is the inverse of $f_{\ast }$ as homomorphisms of abelian groups.