[xsl] XSLT transform to HTML - different output using XMLSpy vs VB6

["Craig Whisenhunt" <craig@xxxxxxxxxxxx>]

I developed a stylesheet using XMLSpy and am now trying to use the XSLT file
within VB6 to transform XML to HTML docs.

Everything works beautifully in XMLSpy but the results of the tranformation
using VB6 are curious (and not acceptable). I'm not sure where to start so
here goes. There are a few issues:

Here's the first part of my stylesheet:
<?xml version="1.0" encoding="UTF-8" ?>
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
xmlns:xs="http://www.w3.org/2001/XMLSchema";
xmlns:fn="http://www.w3.org/2004/07/xpath-functions";
xmlns:xdt="http://www.w3.org/2004/07/xpath-datatypes";>
<xsl:preserve-space elements="*"/>
<xsl:output method="html" encoding="ISO-8859-1" use-character-maps="cm1"
indent="yes"/>
<xsl:output doctype-system="http://www.w3.org/TR/html4/loose.dtd"; />
<xsl:output doctype-public="-//W3C//DTD HTML 4.01//EN" />
<xsl:character-map name="cm1">
<xsl:output-character character="&#160;" string="&amp;nbsp;"/>
</xsl:character-map>

I don't fully understand all these declarations but borrowed them from
another stylesheet example.

The character map is there to force the &nbsp; to be written in the HTML.
One of the main differences in how XMLSpy renders vs VB6 is the &nbsp;

VB6 outputs a blank space while XMLSpy renders a &nbsp; I need the &nbsp;
for formatting and to hold open empty table cells.

Any advice on the declarations at the top of my stylesheet would be
appreciated. Also any knowledge of VB6 in this regard would be helpful.

I'd be happy to post some of the templates but I don't think that is the
problem since XMLSpy renders wonderfullly.

Thanks in advance,
-Craig