Theorem Suppose $X$ and $Y$ are topological spaces, and suppose $f:X\to Y$ is a continuous function. For a subset $A\subset X$ , the restriction of $f$ to $A$ (that is $f|_A$ ) is a continuous mapping $f|_A:A\to Y$ , where $A$ is given the subspace topology from $X$ .

Proof. We need to show that for any open set $V\subset Y$ , we can write $(f|_A)^{-1}(V) = A\cap U$ for some set $U$ that is open in $X$ . However, by the properties of the inverse image (see this page), we have for any open set $V\subset Y$ , $$ (f|_A)^{-1}(V) = A\cap f^{-1}(V).$$ Since $f:X\to Y$ is continuous, $f^{-1}(V)$ is open in $X$ , and our claim follows.