--- a/pnas/pnas.tex	Sun Nov 14 16:00:35 2010 -0800
+++ b/pnas/pnas.tex	Sun Nov 14 16:02:06 2010 -0800
@@ -716,8 +716,8 @@
 
 With this alternate version in hand, it is straightforward to prove the theorem.
 The evaluation map $\Homeo(X)\times BD_j(X)\to BD_j(X)$
-induces a chain map $\CH{X}\ot C_*(BD_j(X))\to C_*(BD_j(X))$
-and hence a map $e_X: \CH{X} \ot \cB\cT_*(X) \to \cB\cT_*(X)$.
+induces a chain map $\CH{X}\tensor C_*(BD_j(X))\to C_*(BD_j(X))$
+and hence a map $e_X: \CH{X} \tensor \cB\cT_*(X) \to \cB\cT_*(X)$.
 It is easy to check that $e_X$ thus defined has the desired properties.
 \end{proof}