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Chapter 10: Problem 6

Show that a map \(f \cdot X \rightarrow Y\) between two topological spaces \(X\) and \(Y\) is contintous if and only if \(f(\bar{U}) \subseteq \overline{f(U)}\) for all sets \(U \subseteq X\) Show that \(f\) is a lwomeomorplism only if \(f(\bar{U})=\overline{f(U)}\) for all sets \(U \subseteq X\)

Short Answer

Expert verified
So the function \(f: X \rightarrow Y\) is continuous if and only if for any subset \(U \subseteq X\), \(f(\bar{U}) \subseteq \overline{f(U)}\). And \(f\) is a homeomorphism if and only if for any subset \(U \subseteq X\), \(f(\bar{U})= \overline{f(U)}\). Therefore, it's been demonstrated that these conditions are both necessary and sufficient for continuity and homeomorphism.

Step by step solution

01

Prove \(f\) is Continuous

Assume that for every set \(U \subseteq X\), \(f(\bar{U}) \subseteq \overline{f(U)}\). Let \(V\) be an arbitrary closed subset of \(Y\). We must prove that \(f^{-1}(V)\) is also closed in \(X\). The complement \(X - f^{-1}(V)\) is an arbitrary subset of \(X\). Using the assumption, we have \(f(\overline{X - f^{-1}(V)}) \subseteq \overline{f(X - f^{-1}(V))}\). As \(f^{-1}(V)\) equals \(X - \overline{X - f^{-1}(V)}\), it follows that \(f^{-1}(V)\) is closed.
02

Show the Bi-conditional for the Continuity of \(f\)

Now we have to show the reverse, i.e., if \(f\) is continuous then for every set \(U \subseteq X\), \(f(\bar{U}) \subseteq \overline{f(U)}\). Assume \(f\) is continuous, and let \(U\) be an arbitrary subset of \(X\). We must prove that \(f(\bar{U}) \subseteq \overline{f(U)}\). Since \(f\) is continuous, \(f(\bar{U})\) is closed in \(Y\), which means \(f(\bar{U}) \subseteq \overline{f(U)}\). Therefore, \(f\) is continuous if and only if \(f(\bar{U}) \subseteq \overline{f(U)}\) for every \(U \subseteq X\).
03

Prove \(f\) is a Homeomorphism

Again, let's assume that for every set \(U \subseteq X\), \(f(\bar{U}) = \overline{f(U)}\). From Step 2, it already follows that \(f\) is continuous. To prove that \(f\) is a homeomorphism, we must show that its inverse \(f^{-1}\) is also continuous. By applying the condition to \(f^{-1}\), we get that \(f^{-1}(\bar{V}) = \overline{f^{-1}(V)}\) for every \(V \subset Y\), thus proving that \(f^{-1}\) is continuous.
04

Show the Bi-conditional for Homeomorphism of \(f\)

For the reverse, assume that \(f\) is a homeomorphism. As a result, both \(f\) and \(f^{-1}\) are continuous (From Steps 1 and 3). Now we must prove that \(f(\bar{U}) = \overline{f(U)}\) for every \(U \subset X\). We already know from Step 2 that \(f(\bar{U}) \subseteq \overline{f(U)}\). For the reverse inclusion, denote \(V = f(U)\). As \(f^{-1}\) is continuous, we have \(f^{-1}(\overline{V}) \subseteq \overline{f^{-1}(V)} = \overline{U}\). As \(f(\overline{U}) \supseteq f(f^{-1}(\overline{V})) = \overline{V}\), we have proved the required.

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Most popular questions from this chapter

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.

Show that a linear map \(T: V \rightarrow W\) between topological vector spaces is continuous everywhere on \(V\) if and only if it is continuous at the origin \(0 \in V\).

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 \(X\)

Show that the following are all norms in the vector space \(\mathbb{R}^{2}\) : $$ \begin{aligned} &\|\mathbf{u}\|_{1}=\sqrt{\left(u_{1}\right)^{2}+\left(u_{2}\right)^{2}} \\ &\|\mathbf{u}\|_{2}=\max \left[\left|u_{1}\right|,\left|u_{2}\right|\right\\} \\\ &\|\mathbf{u}\|_{3}=\left|u_{1}\right|+\left|u_{2}\right| \end{aligned} $$ What are the shapes of the open balls \(B_{a}(\mathrm{u})\) ? Show that the topologes generated by these norms are the same.

If \(G_{0}\) is the component of the identity of a locally connected topological group \(G\), the factor group \(G / G_{0}\) is called the group of components of \(G .\) Show that the group of components is a discrete topological group with respect to the topology induced by the natural projection map \(\pi: g \mapsto g G_{0}\)
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