Show that $f : (X, d_1) \rightarrow (Y, d_2)$ is continuous iff $f^{-1}(U)\in T_1$ for all $U\in T_2$ Let $f : (X, d_1) \rightarrow (Y, d_2)$ be a map between metric spaces and let $T_1$ (and $T_2$ respectively) be the topology on $X$ (and on $Y$ respectively) that is induced by $d_1$ ($d_2$ respectively).
Show using the $\epsilon$-$\delta$ definition of continuity that $f : (X, d_1) \rightarrow (Y, d_2)$ is continuous iff $f^{-1}(U)\in T_1$ for all $U\in T_2$.
$\Rightarrow$" :
Suppose that $f : (X, d_1) \rightarrow (Y, d_2)$ is continuous.
Given $\epsilon > 0$ there exists $\delta > 0$ such that $$d_1(x, y) < \delta \implies d_2(f(x), f(y)) < \epsilon$$ right?
To show that $f^{-1}(U)\in T_1$ for all $U\in T_2$ we have to show that if $a\in f^{-1}(U)$ then $a\in T_1$ ?

Let $f$ is continuous. $U$ be any open set in $Y$. We will show that $f^{-1}(U)$ is open in $X$.  Let $a \in f^{-1}(U)$, $f(a) \in U$. Since $U$ is open there exists $\epsilon>0$ such that $B_Y(f(a),\epsilon)\subseteq U$. Now by continuity at $a$, there exists $\delta>0$ such that $f(B_X(a,\delta))\subseteq B_Y(f(a),\epsilon)\subseteq U$. This implies $a \in B_X(a,\delta)\subseteq f^{-1}(U)$. Thus $f^{-1}(U)$ is open in $X$.
Now let $a \in X$. We will show that $f$ is continuous  at $a$. For any $\epsilon>0$, $f(a)\in B_Y(f(a),\epsilon)$. $f^{-1}(B_Y(f(a),\epsilon))$ is open in $X$ containing $a$. So there exists $\delta>0$ such that $B_X(a, \delta)\subseteq f^{-1}(B_Y(f(a),\epsilon))$. This implies $f(B_X(a, \delta))\subseteq B_Y(f(a),\epsilon)$. Thus $f$ is continuous at $a$.
