Algebraic Topology

1. add Prove that a retract of a Hausdorff space is closed.

   Proof:

   Let $X$ be a Hausdorff space and let $A\subset X$ be a retract of $X$. We will show that $X\setminus A$ is open. Since $A$ is a retract of $X$ there exists a continuous map $f:X\to A$ such that $f\left(x\right)=x$ for all $x\in A$. Notice that $f\left(a\right)\in A$ but $f\left(a\right)\ne a$ since $a\notin A$.

   Since $X$ is Hausdorff there exists disjoint open sets $U$ and $V$ such that $a\in U$ $f\left(a\right)\in V$. It follows that $W=U\cap f^{-1}\left(V\right)$ is an open set containing $a$.

   Now, it suffices to show that $W$ and $A$ are disjoint. Let $y\in W\cap A$. It follows that $f\left(y\right)\ne y$ since $f\left(y\right)\in V$, and for disjoint sets $U$ and $V$, $y\in U$. Hence, $y\notin A$ and $W$ is an open neighborhood of $a$ lying completely in $X\setminus A$.

   Since $a$ was chosen arbitrarily, every point in $X\setminus A$ has an open neighborhood lying completely in $X\setminus A$. Thus, $X\setminus A$ is open and $A$ is closen.

   Q.E.D.

2. add Prove that the product of Hausdorff spaces is Hausdorff.

   Proof:

   Let $X$ and $Y$ be Hausforff topological spaces and define the product topology $X\times Y$. By the definition of product topology, a point in $X\times Y$ is of the form $(x,y)$ where $x\in X$ and $y\in Y$. This means that we can pick two distinct points $(a,b)$ and $(c,d)$ in $X\times Y$ where $a\ne c$ or $b\ne d$ (or both).

   Since $X$ is Hausdorff there exists open disjoint sets $U_1,U_2$ such that $a\in U_1$ and $c\in U_2$. Likewise, since $Y$ is Hausdorff there exists open disjoint sets $V_1,V_2$ such that $b\in V_1$ and $d\in V_2$.

   Without loss of generality${}_1$, assume $b\ne d$. Then $V_1,V_2$ are two open disjoint sets such that $b\in V_1$ and $d\in V_2$. It follows that $X\times V_1$ and $X\times V_2$ are open disjoint sets in $X\times Y$ such that $(a,b)\in X\times V_1$ and $(c,d)\in X\times V_2$. Thus, $X\times Y$ is Hausdorff.

   Q.E.D.