Definition (Reeb Graph). Given a continuous function \(f \colon X \rightarrow \overline{{\mathbb{R}}}\) and \(x \in X\), let \(\pi_f (x)\) be the connected component of \(x\) in \(f^{-1}(f(x))\). In this way we obtain a function \(\pi_f \colon X \rightarrow \pi_f (X) \subset 2^X\) and we endow \(\pi_f (X)\) with the quotient topology^[2]. By the universal property of the quotient space there is a unique continuous function \(\mathcal{R} f \colon \pi_f (X) \rightarrow \overline{{\mathbb{R}}}\) such that \[ \xymatrix{ X \ar[r]^{\pi_f} \ar[dr]_f & \pi_f (X) \ar[d]^{\mathcal{R} f} \\ & \overline{{\mathbb{R}}} } \] commutes, i.e. \(\mathcal{R} f \circ \pi_f = f\).

For another continuous function \(g \colon Y \rightarrow \overline{{\mathbb{R}}}\) and a homomorphism \(\varphi \colon f \rightarrow g\) we may use the universal property of \(\pi_f\) again, to obtain a unique continuous map \(\mathcal{R} \varphi \colon \pi_f (X) \rightarrow \pi_g (Y)\) such that \[ \xymatrix{ X \ar[d]_{\pi_f} \ar[r]^{\varphi} & Y \ar[d]^{\pi_g} \\ \pi_f (X) \ar[r]_{\mathcal{R} \varphi} & \pi_g (Y) } \] commutes, i.e. \(\mathcal{R} \varphi \circ \pi_f = \pi_g \circ \varphi\).

Lemma. Let \(f \colon X \rightarrow \overline{{\mathbb{R}}}\) and \(g \colon Y \rightarrow \overline{{\mathbb{R}}}\) be continuous functions and let \(\varphi \colon f \rightarrow g\) be a homomorphism, then the diagram \[ \xymatrix{ \pi_f (X) \ar@/^/[rr]^{\mathcal{R} \varphi} \ar[dr]_{\mathcal{R} f} & & \pi_g (Y) \ar[dl]^{\mathcal{R} g} \\ & \overline{{\mathbb{R}}} } \] commutes. Or in other words \(\mathcal{R} \varphi\) is a homomorphism from \(\mathcal{R} f\) to \(\mathcal{R} g\) in the category of \(\overline{{\mathbb{R}}}\)-spaces.

Proof. We consider the diagram \[ \xymatrix@C=5pt{ \pi_f (X) \ar[rr]^{\mathcal{R} \varphi} \ar@/_3pc/[ddr]_{\mathcal{R} f} & & \pi_g (Y) \ar@/^3pc/[ddl]^{\mathcal{R} g} \\ X \ar[u]_{\pi_f} \ar[rr]^{\varphi} \ar[dr]^f & & Y \ar[u]^{\pi_g} \ar[dl]_g \\ & \overline{{\mathbb{R}}} . } \] In this diagram the three inner triangles and the square commute. Further \(\pi_f\) and \(\pi_g\) are surjective and thus the outer triangle commutes as well.

Lemma. Let \(f \colon X \rightarrow \overline{{\mathbb{R}}}\), \(g \colon Y \rightarrow \overline{{\mathbb{R}}}\), and \(h \colon Z \rightarrow \overline{{\mathbb{R}}}\) be continuous functions and let \(\varphi \colon f \rightarrow g\) and \(\psi \colon g \rightarrow h\) be two homomorphisms, then \(\mathcal{R} (\psi \circ \varphi) = \mathcal{R} \psi \circ \mathcal{R} \varphi\).

Proof. We consider the diagram \[ \xymatrix{ X \ar[r]^{\varphi} \ar[d]_{\pi_f} & Y \ar[r]^{\psi} \ar[d]_{\pi_g} & Z \ar[d]_{\pi_h} \\ \pi_f (X) \ar[r]_{\mathcal{R} \varphi} & \pi_g (Y) \ar[r]_{\mathcal{R} \psi} & \pi_h (Z) . } \] By definition both inner squares commute, hence the outer square commutes as well. Now it follows from the uniqueness part of the universal property of \(\pi_f\) that \(\mathcal{R} (\psi \circ \varphi) = \mathcal{R} \psi \circ \mathcal{R} \varphi\).