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Let $D_{\phi}$ be the set o discontinuity points of $\phi$. How can i prove that if $g\circ f$ makes sense then $D_{g\circ f} \subset D_{f} \bigcup f^{-1}(D_{g})$?

for $ x \in D_{g\circ f} \Rightarrow x \in D_{f}$ is obvious but what about $f^{-1}(D_{g})$? the exercise do not give any information about $f$ but he use $f^{-1}$, can i assume that $f$ is a bijection?

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The reason for having $x \in f^{-1}(D_g)$ is because $g \circ f$ is defined where $f$ is defined on $D_g$.

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  • $\begingroup$ One-liners rarely make good Answers. While you engaged with one aspect of the problem, much is left to the Reader's imagination to see this as answering the Question as a whole. $\endgroup$ – hardmath Jun 26 '16 at 18:25

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