Compositions of simplicial maps

Redirect to:

• Simplicial maps and chain maps

What happens to the chain maps and the homology maps under compositions?

As one would hope, compositions turn into compositions!

Theorem. The homology map of the composition is the composition of the homology maps.

Let's elaborate and prove this for simplicial maps.

Suppose we have two simplicial maps: $$K {\rightarrow}^f{\rightarrow} L {\rightarrow}^g{\rightarrow} M.$$ These maps induce two chain maps: $$C_k(K) {\rightarrow}^{f_k}{\rightarrow} C_k(L) {\rightarrow}^{g_k}{\rightarrow} C_k(M).$$

It is clear that gf is a simplicial map (exercise). But what is the chain map $(gf)_k$ that it induces?

This is the answer:

Lemma. The chain map of the composition is the composition of the chain maps: $$(gf)_k = g_kf_k.$$

Proof. Suppose we have an n-simplex s: $$s = {\rm conv}\{v_0, v_1, ..., v_n\},$$ where $v_0, v_1, ..., v_n$ are its vertices. To prove the identity, we evaluate its left-hand side and its right-hand side for $s$.

First the right-hand side. By Lemma 2 here, we have either a cloned $k$-simplex in $L$ with $k+1$ vertices or a collapsed, lower dimensional, simplex:

• Case 1: $(gf)_k(s) = {\rm conv}\{(gf)(v_0), ..., (gf)(v_n) \}$ if $(gf)(v_i) {\neq} (gf)(v_j)$ for all $i{\neq}j$, or
• Case 2: $(gf)_k(s) = 0$ otherwise.

Now the left-hand side. Invoke the lemma for $f$:

• Case 1: $f_k(s) = {\rm conv}\{f(v_0), ..., f(v_n) \}$ if $f(v_i) {\neq} f(v_j)$ for all $i{\neq}j$, or
• Case 2: $f_k(s) = 0$ otherwise.

Apply $g$ to the simplex above:

• Case 1: $g_k(f_k(s)) = g_k({\rm conv}\{f(v_0), ..., f(v_n)\})$ if $f(v_i) {\neq} f(v_j)$ for all $i{\neq}j$, or
• Case 2: $g_k(f_k(s)) = 0$ otherwise.

Now invoke the lemma, applied to Case 1. Then there are three cases:

• Case 1a: $g_k(f_k(s)) = {\rm conv}\{g(f(v_0)), ..., g(f(v_n)) \}$ if $f(v_i) {\neq} f(v_j)$ and $g(f(v_i)) {\neq} g(f(v_j))$ for all $i{\neq}j$, or
• Case 1b: $g_k(f_k(s)) = 0$ if $f(v_i) {\neq} f(v_j)$ but $g(f(v_i)) = g(f(v_j))$ for some $i{\neq}j$, or
• Case 2: $g_k(f_k(s)) = 0$ if $f(v_i) = f(v_j)$ for some $i{\neq}j$.

In Case 1, the second condition implies the first. Meanwhile Case 1b and Case 2 can be merged. So,

• Case 1a: $g_k(f_k(s)) = {\rm conv}\{g(f(v_0)), ..., g(f(v_n))\}$ if $g(f(v_i)) {\neq} g(f(v_j))$ for all $i{\neq}j$, or
• Cases 1b2: $g_k(f_k(s)) = 0$ if $g(f(v_i)) = g(f(v_j))$ for some $i{\neq}j$.

Finally we observe that Case 1 = Case 1a and Case 2 = Case 1b2, because $$(gf)_k(s) = g_k(f_k(s)). $$ $\blacksquare$

Theorem. The homology map of the composition is the composition of the homology maps: $$(gf)_* = g_*f_*.$$

Proof. Exercise. $\blacksquare$