Show that h is a continuous function from X to Z where Z is endowed with the subspace topology.

Question

Let f be a continuous function from a topological space X to another topological space Y. Let Z be a subset of Y which contains f(X) (the image of f). Define a function h from X to Z by putting h(x) = f(x) for each x ∈ X. Show that h is a continuous function from X to Z where Z is endowed with the subspace topology.

Answer 1

Let $V$ be an open set in $Z$. By the definition of the subspace topology, $V=Z \cap U$, where $U$ is an open set in $Y$. Now

$$\begin{align} f^{-1}(Z \cap U) & = f^{-1}(Z) \cap f^{-1}(U) \\ & = X \cap f^{-1}(U) \\ & = f^{-1}(U) \end{align} $$

Because the preimage commutes with intersections and unions, and because $f^{-1}(Z)=X$, by the assumption that $f(X) \subseteq Z$. As $f^{-1}(U)$ is open by assumption, the result follows.

Remark: Really all that's going on here is applying set theory relations and the definition of the subspace topology.