We study the composition operator \documentclass[12pt]{minimal}
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\begin{document}$$T_f(g):= f\circ g$$\end{document} on Besov spaces \documentclass[12pt]{minimal}
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\begin{document}$$B_{{p},{q}}^{s}(\mathbb{R })$$\end{document}. In case \documentclass[12pt]{minimal}
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\begin{document}$$1 < p< +\infty ,\, 0< q \le +\infty $$\end{document} and \documentclass[12pt]{minimal}
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\begin{document}$$s>1+ (1/p)$$\end{document}, we will prove that the operator \documentclass[12pt]{minimal}
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\begin{document}$$T_f$$\end{document} maps \documentclass[12pt]{minimal}
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\begin{document}$$B_{{p},{q}}^{s}(\mathbb{R })$$\end{document} to itself if, and only if, \documentclass[12pt]{minimal}
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\begin{document}$$f(0)=0$$\end{document} and \documentclass[12pt]{minimal}
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\begin{document}$$f$$\end{document} belongs locally to \documentclass[12pt]{minimal}
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\begin{document}$$B_{{p},{q}}^{s}(\mathbb{R })$$\end{document}. For the case \documentclass[12pt]{minimal}
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\begin{document}$$p=q$$\end{document}, i.e., in case of Slobodeckij spaces, we can extend our results from the real line to \documentclass[12pt]{minimal}
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\begin{document}$$\mathbb{R }^n$$\end{document}.