Question

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

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

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.

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.

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.

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.

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.

Key concepts

Continuous Map

In topology, a continuous map refers to a function between two topological spaces that preserves the structure of the spaces. This concept is like stretching or bending without tearing or gluing. For a map \( f: X \to Y \), this means if you have an open set in \( Y \), the preimage (or the set of all points in \( X \) that map to this open set) under \( f \) needs to be open in \( X \).

One way to think about continuous maps is through the lens of neighborhood preservation; points close together in \( X \) must map to points close together in \( Y \).

In our exercise, the map \( f: X \to Y \) is continuous, which is important because it ensures that when we later explore the graph and its properties, the continuity allows for the necessary topological configurations.

Homeomorphism

A homeomorphism is a very strong type of continuous map. It's essentially a 'topological isomorphism.' This means, if two spaces are homeomorphic, they are topologically the same, even if they might look different at a first glance.

To say that a map \( h: G \to X \) (as defined in the exercise) is a homeomorphism, we need it to satisfy two conditions:

• Bijective: \( h \) pairs every element in \( G \) with a unique element in \( X \), and vice versa.
• Both \( h \) and its inverse \( h^{-1} \) are continuous. This preserves the topological properties in both directions.

Demonstrating these conditions ensures that \( G \) retains the open set structure of \( X \) and thus, is homeomorphic to it, confirming their equivalence in topological terms.

Relative Topology

Relative topology is a concept where we consider a subset of a topological space, but with the 'inherited' open sets from the larger space. Imagine a big cake and taking a slice— the slice inherits the layers and flavors from the whole cake but is now considered as its own object.

Given the subset \( G \) from the product space \( X \times Y \), the open sets in \( G \) are precisely the intersections of \( G \) with open sets in \( X \times Y \). This is crucial when we discussed continuity in our exercise because any proofs about openness or continuity must respect this inherited structure.

Product Topology

Product topology is a way of constructing a new topological space from two existing ones. It involves creating a 'product' such that the resulting space has elements that are ordered pairs from its original spaces.

For spaces \( X \) and \( Y \), the product topology on \( X \times Y \) consists of open sets that are products of open sets from each of \( X \) and \( Y \). These might look like \( U \times V \), where \( U \) is open in \( X \) and \( V \) is open in \( Y \).

In the continuous map exercise, understanding the product topology helps in visualizing how \( G \) fits within the larger environment of \( X \times Y \) and why relative topology could be applied here to show homeomorphism with \( X \). That's because the openness in the product space checks for each component independently, which aligns nicely with the function \( f \).