Functions with Compact Support and Algebraic Structures

Functions with Compact SupportRing‐valued functionsModule‐valued functionsExample: infinite coproduct

 

Functions with Compact Support

Let $G$$G$ be a group and $X$$X$ a topological space. Then ${\mathrm{Hom}}_{\mathbf{Set}}(X,G)$$\mathrm{Hom}_{\mathbf{Set}}(X,G)$ is a group under pointwise multiplication:

$$(fg)(x)=f(x)g(x),f^{-1}(x)=f(x{)}^{-1},e(x)=e.$$

Define

$$D(f)=\{x\in X:f(x)\ne e\},\mathrm{supp}(f)=\overline{D(f)}.$$

Lemma 1. $\overline{U\cup V}=\bar{U}\cup \bar{V}.$$\overline{U\cup V} = \overline{U}\,\cup\,\overline{V}.$

Proof. It is easy to see that

$$\bar{U}\subseteq K⟺U\subseteq i(K)$$

The closure operator $\overline{(-)}$$\overline{(-)}$ is left adjoint to the inclusion of closed sets into all subsets, hence preserves unions (colimits). $◻$$\square$

It follows that

$$\mathrm{supp}(fg)=\overline{D(fg)}\subseteq \overline{D(f)\cup D(g)}=\mathrm{supp}(f)\cup \mathrm{supp}(g).$$

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Lemma 2. The union of two compact subsets of $X$$X$ is compact.

Proof. If $\{U_i{\}}_{i\in I}$$\{U_i\}_{i\in I}$ covers $U\cup V$$U\cup V$, it covers each of $U$$U$ and $V$$V$. By compactness extract finite subcovers for $U$$U$ and $V$$V$, then take their union.
$◻$$\square$

Lemma 3. A closed subset of a compact space is compact.

Proof. If $\{U_i\}$$\{U_i\}$ covers $F\subseteq K$$F\subseteq K$, then $\{X\setminus F\}\cup \{U_i\}$$\{X\setminus F\}\cup\{U_i\}$ covers $K$$K$. Extract a finite subcover and discard $X\setminus F$$X\setminus F$.
$◻$$\square$

Corollary. If $f,g$$f,g$ have compact support then so does $fg$$f g$.

Proposition.

$$K=\{f\in {\mathrm{Hom}}_{\mathbf{Set}}(X,G):\mathrm{supp}(f)\text{ is compact}\}$$

is a subgroup.

Proof. The constant map $e$$e$ has empty (hence compact) support. Closure under products follows from the corollary. For inverses note $\mathrm{supp}(f^{-1})=\mathrm{supp}(f)$$\mathrm{supp}(f^{-1})=\mathrm{supp}(f)$.
$◻$$\square$

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Ring‐valued functions

If $R$$R$ is a ring, pointwise addition and multiplication make ${\mathrm{Hom}}_{\mathbf{Set}}(X,R)$$\mathrm{Hom}_{\mathbf{Set}}(X,R)$ into a ring. Defining $D(f)$$D(f)$ and $\mathrm{supp}(f)$$\mathrm{supp}(f)$ as above, one shows:

• $\mathrm{supp}(f+g)\subseteq \mathrm{supp}(f)\cup \mathrm{supp}(g)$$\mathrm{supp}(f+g)\subseteq \mathrm{supp}(f)\cup\mathrm{supp}(g)$.

• $\mathrm{supp}(fg)\subseteq \mathrm{supp}(f)\cap \mathrm{supp}(g)$$\mathrm{supp}(f g)\subseteq \mathrm{supp}(f)\cap\mathrm{supp}(g)$.

Hence

$$K=\{f:X\to R:\mathrm{supp}(f)\text{ is compact}\}$$

is an ideal, and if $X$$X$ itself is compact, a subring.

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Module‐valued functions

If $M$$M$ is an $R$$R$‐module then ${\mathrm{Hom}}_{\mathbf{Set}}(X,M)$$\mathrm{Hom}_{\mathbf{Set}}(X,M)$ is an $R$$R$‐module pointwise, and

$$\mathrm{supp}(f+g)\subseteq \mathrm{supp}(f)\cup \mathrm{supp}(g),\mathrm{supp}(rf)\subseteq \mathrm{supp}(f).$$

Thus

$$K=\{f:X\to M:\mathrm{supp}(f)\text{ is compact}\}$$

is an $R$$R$‐submodule.

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Example: infinite coproduct

Let $I$$I$ be discrete. Then the only compact subsets of $I$$I$ are the finite ones, and we could construct the coproduct as:

$$\underset{i\in I}{⨁}{M}_i:=\{(x_i{)}_{i\in I}\in \prod _{i\in I}{M}_i|(x_i{)}_{i\in I}\text{ has compact support}\}$$