Prove that \(\mathbb R\smallsetminus \{0\}\to \mathbb R\text{,}\)\(x\mapsto 1/x\) is continuous but admits no continuous extension to a map with domain \(\mathbb R\text{.}\) (Part of your task is to make the meaning of this prompt precise. As much as possible, use "preimage of an open is open" language in your proof, as opposed to \(\varepsilon\)-\(\delta\) language.)
Rigorously prove that \(\mathrm{SO}(2)\) is homeomorphic to \(S^1\text{.}\) Here \(\mathrm{SO}(2) = \mathrm{SO}_2(\mathbb R)\) the set of \(2\times 2\) real orthogonal matrices with determinant \(1\) given the subspace topology inside \(\mathbb R^{2\times 2}\text{,}\) and \(S^1\) is the unit circle given the subspace topology inside \(\mathbb R^2\text{.}\) You may freely use the fact that every element of \(\mathrm{SO}(2)\) is of the form \(\begin{pmatrix}
\cos \theta \amp -\sin \theta\\
\sin \theta \amp \cos \theta
\end{pmatrix}\) for \(\theta\) some real number.
Consider a morphism \(f\colon x\to y\) in a category \(\mathsf C\text{.}\) Prove the following statement:
if there exist morphisms \(g,h\colon y\rightrightarrows x\) such that \(gf=\operatorname{id}_x\) and \(gf=\operatorname{id}_y\text{,}\) then \(g=h\) and \(f\) is an isomorphism.
Prove that functors preserve isomorphisms, i.e., if \(F\colon \mathsf C\to \mathsf D\) is a functor and \(f\) is an isomorhism in \(\mathsf C\text{,}\) then \(Ff\) is an isomorphism in \(\mathsf D\text{.}\)
Give an example of a functor that does not refelct isomorphisms; that is, find categories \(\mathsf C\text{,}\)\(\mathsf D\text{,}\) a functor \(F\colon \mathsf C\to \mathsf D\text{,}\) and a morphism \(f\) of \(\mathsf C\) for which \(Ff\) is an isomorphism but \(f\) is not.