Question

Let X and Y be two topological spaces with Y Hausdorff, and let f,g : X → Y be two continuous maps. Show that: (a) If f is one-to-one, then X is Hausdorff; (b) If f(x) = g(x) ∀x ∈ A, where ¯A = X, then f = g; (c) The graph Γ = { (x,f(x)) | x ∈ X} is closed in X ×Y .

Answer 1

$a)\ \text{We take two elements } x_1\text{ and } x_2\in X, x_1\neq x_2.\\ \text{Then we find }f(x_1) \text{ and } f(x_2), f(x_1)\neq f(x_2).\\ \text{As } Y \text{ is Hausdorff there exist two neighbourhoods }U\text{ and }\\ V\text{ in Y}: f(x_1)\subset U,\ f(x_2)\subset V,\ U\cap V=\empty.\\ \empty=f^{-1}(U\cap V)=f^{-1}(U)\cap f^{-1}(V).\\ \text{We found two neighbourhoods }f^{-1}(U)\text{ and }f^{-1}(V)\\ \text{of } x_1 \text{ and } x_2\text{ respectively such that }f^{-1}(U)\cap f^{-1}(V)=\empty.\\ X\text{ is Hausdorff}.\\ b)\text{ We should show that } f(x)=g(x)\text{ on the set of all the boundary points of }X.\\ \text{Let }x_0\text{ be a boundary point of }X.\\ \text{We reach }x_0\text{ from the side of }X:\\ \lim_{x\rightarrow x_0}f(x)=f(x_0)\text{ (f is continuous)}.\\ \lim_{x\rightarrow x_0}g(x)=g(x_0)\text{ (g is continuous)}.\\ f(x)=g(x)\text{ for all the interior points. So }f(x_0)=g(x_0).$

c) We take a point $(x,y)\notin \Gamma.$ Because $y\neq f(x)$ and $Y$ is Hausdorff, there exist disjoint open sets $V,\ V^{\prime}$ with $f(x)\in V$ and $y\in V^{\prime}$. $f$ is continuous, so $U=f^{-1}(V)$ is open with $x\in U$. We see that $U\times V^{\prime}$ is an open neighborhood of $(x,y)$ that does not intersect the graph. Indeed, if we suppose (for contradiction) that $(z,f(z))\in U\times V^{\prime}$, then it would follow that $f(z)\in V^{\prime}$. However, $z\in U=f^{-1}(V)$ which means that $f(z)\in V$. Thus, $f(z)∈V\cap V^{\prime}$ contradicting the fact that $V,\ V^{\prime}$ are disjoint.