<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet
    version="1.0"
    xmlns="http://www.icpsr.umich.edu/DDI"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    >

    <xsl:template match="codeBook">
       <xsl:copy>
         <xsl:apply-templates select="@*"/>
        <docDscr>
           ...
               <titl>
                 <xsl:value-of select="stdyDscr/citation/titlStmt/titl"/>
               </titl>
           ...
         </docDscr>
         <xsl:apply-templates select="*"/>
        </xsl:copy>
    </xsl:template>

    ...
</xsl:stylesheet>

<?xml version="1.0" encoding="utf-8"?>
<codeBook xmlns="http://www.icpsr.umich.edu/DDI" >
    <stdyDscr>
	<citation>
            <titlStmt>
                <titl>foo</titl>
            </titleStmt>
        </citation>
    </stdyDscr>
</codeBook>

I thought to myself, there should be some way I can match the default namespace in the source document?

there is nothing special about the default namespace just think of it as
having prefix "" with a special syntax trick that if teh prefix is ""
you don't use a colon.

Xpath is essentially blind to the prefixes used in the source, it
doesn't matter if it is <x:foo> or <b:foo> or <foo> so long as in each
case it is foo in the same namespace you always use the same Xpath to
select the element.

David

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