Problem 1

Exercise 7.1 (Continuous bijections, compactness, Hausdorff, and gluing)  

1. Let $f : X \to Y$ be a continuous bijection where $X$ is compact and $Y$ is Hausdorff. Then $f$ is a homeomorphism.

2. Let $X=A\cup B$ where $A$ and $B$ are closed subsets of $X$. Suppose $f : X \to Y$ is a map such that $f\vert _A$ and $f\vert _B$ are continuous. Then $f$ is continuous.

Proof. For part (a): we show $(f^{-1})^{-1}(S)$ is closed for any closed $S\subseteq X$. Let $S\subseteq X$ be closed. The inverses simplify to $f(S)$ because $f$ is a bijection. Since $X$ is compact and $S$ is closed, it follows that $S$ is compact. Continuous images of compact maps are compact, so we know $f(S)$ is compact. Since $Y$ is Hausdorff, this implies $f(S)$ is closed.

For part (b), let us glue by hand. We know $f(X)=f(A)\cup f(B)$, so if $U\subseteq f(X)$ is open, $U=U\cap f(A) \cup U\cap f(B)$. By writing this we see

$$f^{-1}(U) = f^{-1}(U\cap f(A)\cup U\cap f(B)) = f^{-1}(U\cap f(A)) \cup f^{-1}(U\cap f(B))$$ (7.1)

$$= (f\vert _A)^{-1}(U)\cup (f\vert _B)^{-1}(U)$$ (7.2)

Since these restrictions are continuous, $f^{-1}(U)$ is a union of open sets, therefore indicating that $f$ is continuous. $\qedsymbol$