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Chapter 10: Problem 13

Let \(X\) and \(Y\) be topological spaccs and \(f . X \times Y \rightarrow X\) a continuous map. For each fixed \(a \in X\) show that the map \(f_{n}: Y \rightarrow X\) defined by \(f_{\theta}(v)=f(a, v)\) is contunous.

Short Answer

Expert verified
Yes, with each fixed \(a \in X\), the map \(f_{\theta}: Y \rightarrow X\) defined by \(f_{\theta}(v)=f(a, v)\) is continuous.

Step by step solution

01

Define the function

The first step is to define the function \(f_{\theta}: Y \rightarrow X\), which is given by \(f_{\theta}(v) = f(a, v)\) where \(a\) is a fixed point in \(X\). By definition, for each \(a \in X\), we have a function \(f_{\theta}: Y \rightarrow X\).
02

Check the pre-image of an open subset

The next step is to check for any open subset \(U\) of \(X\), the pre-image under \(f_{\theta}\), i.e., \(f_{\theta}^{-1}(U)\) is open in \(Y\). Here \(f_{\theta}^{-1}(U) = \{v \in Y : f_{\theta}(v) \in U \}\).
03

Verify whether the pre-image is open

Observe that \(f_{\theta}^{-1}(U) = \{v \in Y : f_{\theta}(v) \in U \} = \{v \in Y : f(a, v) \in U \}\). This set is actually equal to \(f^{-1}(U \times Y)\), where \(f: X \times Y \rightarrow X\) is the original continuous map. Since \(f\) is continuous and \(U \times Y\) is open in \(X \times Y\), \(f^{-1}(U \times Y)\) is open in \(X \times Y\). Hence, \(f_{\theta}^{-1}(U)\) which is a subset of \(f^{-1}(U \times Y)\) is open in \(Y\). Hence \(f_{\theta}: Y \rightarrow X\) is continuous.

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