Proving the inverse of a continuous function is also continuous

Solution 1:

For proving it we use the if and only if condition that continuous function bring back closed sets to closed sets.

Let $F$ be a closed set in $E$. Since $E$ is compact, $F$ is compact. hence $f(F)$ is compact and hence closed.

Now $(f^{-1})^{-1}(F)=f(F)$ is closed.

Since the choice of $F$ was arbitrary, $(f^{-1})$ brings back closed sets to closed sets. hence $f^{-1}$ is continuous.

Solution 2:

We need to prove that $f(A)$ is closed for all closed set $A$ in compact space $K$.

Indeed, $\forall \{y_n \} \subset f(A)$, we assume that $y_n \to y$. We will show that $y \in f(A)$.

We have $f$ is continuous and A is a compact set, so $f(A)$ is also a compact set. By the definition of compact set, we obtain that $\forall \{ y_n\} \subset f(A)$, $\exists \{y_{n_k} \} \subset \{ y_n\}$: $y_{n_k}\to y_0 \in f(A)$. Since $y_n \to y$, $y_{n_k}\to y$. Thus $y=y_0\in f(A)$.