Representable functor in Calculus

Let us define a functor $T:\mathrm{Top}\to \mathrm{Set}$$T:\mathrm{Top}\to\mathrm{Set}$ as follows:

For objects, we have $T(X):=\{f\in C(\mathbb{R},X)\mid f(x+T)=f(x)\}$$T(X):=\{f\in C(\mathbb R,X)\mid f(x+T)=f(x)\}$. For a continuous function $g:X\to Y$$g:X\to Y$, we define $f⟼g\circ f\in T(Y)$$f\longmapsto g\circ f\in T(Y)$.

This functor is representable; we have $T(-)\cong {\mathrm{Hom}}_{\mathrm{Top}}(S^1,-)$$T(-)\cong\mathrm{Hom}_{\mathrm{Top}}(S^1,-)$.

Let us define a functor $F:\mathrm{Haus}\to \mathrm{Set}$$F:\mathrm{Haus}\to\mathrm{Set}$ as follows:

For objects, $F$$F$ maps $X$$X$ to all the convergent sequences on $X$$X$. For a continuous function $g:X\to Y$$g:X\to Y$, we know it preserves convergence. This functor is representable as well, though not as obviously as the functor $T$$T$ above.

Lemma. Let $f:\mathbb{N}\to X$$f:\mathbb N\to X$ be a sequence. Then $f$$f$ is convergent if and only if $\hat{f}:\mathbb{N}\cup \{\mathrm{\infty }\}\to X$$\hat f:\mathbb N\cup\{\infty\}\to X$ is continuous, where $\mathbb{N}\cup \{\mathrm{\infty }\}$$\mathbb N\cup\{\infty\}$ is the one-point compactification of $\mathbb{N}$$\mathbb N$ with the discrete topology.

Proof.

Notice that in $\mathbb{N}\cup \{\mathrm{\infty }\}$$\mathbb N\cup\{\infty\}$, the open sets are exactly all subsets of $\mathbb{N}$$\mathbb N$ and sets of the form $(\mathbb{N}-C)\cup \{\mathrm{\infty }\}$$(\mathbb N - C)\cup \{\infty\}$, where $C$$C$ is closed and compact in $\mathbb{N}$$\mathbb N$, i.e., a finite subset of $\mathbb{N}$$\mathbb N$. Notice that $\hat{f}$$\hat f$ is automatically continuous at points in $\mathbb{N}$$\mathbb N$, so we only need to prove that:

$f$$f$ is convergent $⟺$$\iff$ $\hat{f}$$\hat f$ is continuous at $\mathrm{\infty }$$\infty$.

Assume $\hat{f}$$\hat f$ is continuous at $\mathrm{\infty }$$\infty$. Then, for every open neighborhood of $\hat{f}(\mathrm{\infty })$$\hat f(\infty)$, its preimage under $\hat{f}$$\hat f$ will be an open neighborhood of $\mathrm{\infty }$$\infty$.

Such a neighborhood has the form $\mathbb{N}$$\mathbb N$ minus a finite subset. For each open set $U$$U$ with $\hat{f}(\mathrm{\infty })\in U$$\hat f(\infty)\in U$, there exists an $N$$N$ such that for all $n\ge N$$n\ge N$, $f(n)\in U$$f(n)\in U$. Hence, $f$$f$ is convergent. One might consider the sequence $a_n=n$$a_n=n$ converging to $\mathrm{\infty }$$\infty$, so the sequence $\hat{f}(a_n)$$\hat f(a_n)$ should be convergent as well.

Conversely, assume $f$$f$ converges to $x$$x$, and define $\hat{f}(\mathrm{\infty })=x$$\hat f(\infty)=x$. Then, by definition, for each open set $U$$U$ containing $x$$x$, there exists an $N\in \mathbb{N}$$N\in\mathbb N$ such that for all $n\ge N$$n\ge N$, we have $\hat{f}(n)\in U$$\hat f(n)\in U$. Then the preimage of $U$$U$ under $\hat{f}$$\hat f$ is precisely:

$$\{\mathrm{\infty }\}\cup (\mathbb{N}-\{0,\dots ,N-1\})\cup \{n\in \{0,\dots ,N-1\}\mid f(n)\in U\}$$

,which is open. Hence, we have shown that $\hat{f}$$\hat f$ is continuous. $◻$$\square$

Now we have another viewpoint on the fact that continuous functions map convergent sequences to convergent sequences; it follows directly from the fact that the composition of continuous functions is continuous.

Corollary. The functor $F$$F$ is representable. Specifically, we have:

$$F\cong {\mathrm{Hom}}_{\mathrm{Haus}}({\mathbb{N}}^{\ast },-),$$

where ${\mathbb{N}}^{\ast }$$\mathbb N^*$ is the one-point compactification of $\mathbb{N}$$\mathbb N$.