Exercise · ZW-CONTINUOUS-MAP-TOPOLOGICAL-007 · ★★☆ Intermediate
If \(f:X\to Y\) and \(g:Y\to Z\) are continuous, what can be said about \(g\circ f\)?
Choose one option.
A It is continuous only if \(f\) is bijective B It is continuous only if \(g\) is injective C It need not be continuous D It is continuous, since \((g\circ f)^{-1}(W)=f^{-1}\big(g^{-1}(W)\big)\)
Show answer D. It is continuous, since \((g\circ f)^{-1}(W)=f^{-1}\big(g^{-1}(W)\big)\)
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Explanation If \(W\) is open in \(Z\), \(g^{-1}(W)\) is open in \(Y\), and then \(f^{-1}(g^{-1}(W))\) is open in \(X\). No extra hypothesis is needed.
A function \(f:X\to Y\) between topological spaces \((X,\mathcal{T}_X)\) and \((Y,\mathcal{T}_Y)\) is continuous if the preimage of every open set \(V\) of \(Y\) is an open set of \(X\).
\[V\in\mathcal{T}_Y\ \Longrightarrow\ f^{-1}(V)\in\mathcal{T}_X\]
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