# FA1

Let $X\,$ and $Y\,$ be metric spaces, and $f:X\to Y\,$ be a mapping.

(i) Prove that if $f^{{-1}}(G)\,$ is open whenever $G\subset Y\,$ is open, then $f\,$ is continuous.

(ii) Prove that $f\,$ is continuous if and only if $f^{{-1}}(F)\,$ is closed whenever $F\subset Y\,$ is closed.

## (i)

Suppose $f^{{-1}}(G)\,$ is open whenever $G\subset Y\,$ is open. Let $x_{0}\in X\,$ and $\epsilon >0\,$. Since the ball $B(f(x_{0}),\epsilon )\,$ is open, so is $f^{{-1}}\left[B(f(x_{0}),\epsilon ))\right]\,$. Since $f(x_{0})\in B(f(x_{0}),\epsilon ),x_{0}\in f^{{-1}}\left[B(f(x_{0}),\epsilon )\right]\,$. Since $f^{{-1}}\left[B(f(x_{0}),\epsilon )\right]$ is open there exists a $\delta >0\,$ such that $B(x_{0},\delta )\subset f^{{-1}}\left[B(f(x_{0}),\epsilon )\right]\,$. This implies $f(B(x_{0},\delta ))\subset B(f(x_{0}),\epsilon )\,$ making $f\,$ continuous at $x_{0}\,$. Since $x_{0}\,$ was arbitrary, $f\,$ is continuous.

## (ii)

Suppose $f^{{-1}}(F)\,$ is closed whenever $F\in Y\,$ is closed. Let $G\subset Y\,$ be open. Then $G^{c}\,$ is closed. By the hypothesis, $f^{{-1}}(G^{c})\,$ is closed.

If $x\in f^{{-1}}(G^{c})\,$, then $f(x)\in G^{c},f(x)\notin G,x\notin f^{{-1}}(G)\,$ and $x\in \left[f^{{-1}}(G)\right]^{c}\,$. Similarly, if $x\in \left[f^{{-1}}(G)\right]^{c}\,$ then $x\notin f^{{-1}}(G),f(x)\notin G,f(x)\in G^{c}\,$, and $x\in f^{{-1}}(G^{c})\,$. Therefore $f^{{-1}}(G^{c})=\left[f^{{-1}}(G)\right]^{c}\,$.

Now, $f^{{-1}}(G^{c})=\left[f^{{-1}}(G)\right]^{c}\,$ is closed, so $f^{{-1}}(G)\,$ is open. So $f^{{-1}}(G)\,$ is open whenever $G\subset Y\,$ is open. From part (i), $f\,$ is continuous.

Suppose $f\,$ is continuous. Let $G\subset Y\,$ be closed. Let $\{x_{n}\}\subset f^{{-1}}(G)\,$ converge to $x\in X\,$. Since $f\,$ is continuous $\{f(x_{n})\}\subset G\,$ converges to $f(x)\in Y\,$. Since $G\,$ is closed, $f(x)\in G\,$ and so $x\in f^{{-1}}(G)\,$ which means $f^{{-1}}(G)\,$ is closed (since this is the limit of the arbitrary sequence chosen above). Therefore $f^{{-1}}(G)\,$ is closed whenever $G\,$ is closed.