Past exam of the options II course of the University of Oxford / 2005 / E

The product topology on $X\times Y$ has the open rectangles $U\times V$, with $U$ open in $X$ and $V$ open in $Y$, as a topological basis. The projection preimages are $p_X^{-1}(U)=U\times Y$ and $p_Y^{-1}(V)=X\times V$, so both projections are continuous.

If $f:Z\to X\times Y$ is continuous, both coordinate compositions are continuous. Conversely, if both are continuous, thenis open for every basic rectangle, proving continuity of $f$. Given paths between two chosen endpoints in each factor, their coordinatewise pair is therefore a continuous path in the product, proving that a product of path-connected spaces is path-connected.

For the compact-neighborhood assertion, fix $x\in K$. For each $y\in L$, choose an open rectangle $U_y\times V_y$ containing $(x,y)$ inside $W$. Compactness of $L$ chooses finitely many $V_{y_j}$ covering $L$. PutThen $x\in U_x$, $L\subseteq V_x$, and $U_x\times V_x\subseteq W$: any point of $V_x$ belongs to one of those rectangles' second factors, while $U_x$ lies in every first factor.

Compactness of $K$ now selects finitely many $U_{x_i}$ covering $K$. SetThese are open, contain $K$ and $L$, and if $(u,v)\in U\times V$, choose an $i$ with $u\in U_{x_i}$ and use $v\in V_{x_i}$. HenceIf $K$ or $L$ is empty, take the corresponding open factor empty and the other factor the whole space; the assertions then hold directly. No Hausdorff assumption on the factors is needed.

A Hausdorff space separates any two distinct points by disjoint open neighborhoods. For a subspace $A\subseteq X$, intersect such neighborhoods in $X$ with $A$ to obtain disjoint neighborhoods in the subspace topology. Thus subspaces inherit the Hausdorff property.

If $(z_1,z_2)$ lies off the diagonal of a Hausdorff space's square, choose disjoint open neighborhoods $U$ of $z_1$ and $V$ of $z_2$. The open rectangle $U\times V$ contains the pair and avoids the diagonal. Therefore its complement is open and the diagonal is closed.

The map $F:X\times Y\to Z\times Z$ given by $F(x,y)=(f(x),g(y))$ is continuous by the coordinate criterion from E3. The equality locus is $F^{-1}({\Delta }_Z)$ and is closed as an inverse image of a closed set.

Let $q:X\to X/\sim$ be the quotient map. If the quotient is Hausdorff, apply the same argument to $q$ in both coordinates:is closed in $X\times X$.

Conversely suppose $R$ is closed and $q$ is open. Take two distinct quotient points, with representatives $x,y$. Since $(x,y)\notin R$, an open rectangle $U\times V$ around it avoids $R$. The images $q(U)$ and $q(V)$ are open quotient neighborhoods. They are disjoint, because a quotient point in their intersection would have equivalent representatives $u\in U$, $v\in V$, giving $(u,v)\in R$. Hence the quotient is Hausdorff. Thus closedness of the relation and openness of the quotient map suffice for a Hausdorff quotient.