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Chapter 10: Problem 12

If \(f: X \rightarrow Y\) is a continuous map betwecn lopological spaces, we define its graph to be the set \(G=\\{(x, f(x)) \mid x \in, X] \subseteq X \times Y\). Show that if \(G\) is given the relative topology induced by the topological product \(X \times Y\) then it is homeonorphic to the lopological space \(X\)

Short Answer

Expert verified
The graph \(G\) of the continuous map \(f: X \to Y\) is homeomorphic to the topological space \(X\) when \(G\) is equipped with the relative topology induced by the topological product \(X \times Y\). This is due to the existence of a continuous bijective map \(h: G \to X\) and its continuous inverse.

Step by step solution

01

Understand the Problem Statement

We're given a continuous map \(f: X \to Y\) between two topological spaces. The graph \(G = \{(x, f(x)) \mid x \in X\}\) of this map is a subset of the product space \(X \times Y\). We're required to show that, when \(G\) is equipped with the relative topology from \(X \times Y\), it is homeomorphic to \(X\). This essentially means that there exists a continuous bijective map from \(G\) to \(X\) with continuous inverse.
02

Define the Homeomorphism

Define the function \(h: G \to X\) by \(h(x, f(x)) = x\). Clearly, \(h\) is a bijection. The task now is to show that both \(h\) and its inverse \(h^{-1}\) are continuous functions.
03

Prove that \(h\) is continuous

We show that the preimage of any open set under \(h\) is open in \(G\). Let \(U\) be an open set in \(X\). Then \(h^{-1}(U) = U \times f(U)\), which is open in \(G\) as \(f\) is continuous and \(G\) is given the relative topology induced by \(X \times Y\). Thus, \(h\) is continuous.
04

Prove that \(h^{-1}\) is continuous

We now prove that \(h^{-1}: X \to G\) defined by \(h^{-1}(x) = (x, f(x))\) is continuous. Notice that the map is well-defined as \((x, f(x)) \in G\) for every \(x \in X\). Now, given an open set \(V \subseteq G\), the set \(h(V) = \{x \mid (x, f(x)) \in V\}\) is open in \(X\), since \(f\) is continuous. Hence \(h^{-1}\) is continuous as well.
05

Conclude the Proof

we've constructed a continuous bijective map \(h: G \to X\) with a continuous inverse - by definition, this means \(G\) is homeomorphic to \(X\). This completes the proof.

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