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Deserialize object

suggest change 
public Foo DeserializeFoo(string fileName)
{
    var serializer = new XmlSerializer(typeof(Foo));
    using (var stream = File.OpenRead(fileName))
    {
        return (Foo)serializer.Deserialize(stream);
    }
}
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XmlSerializer:
* XmlSerializer
* Formatting Custom DateTime format
* Serialize object
* Deserialize object
* Behaviour Map array name to property XmlArray
* Behaviour Map Element name to Property
* Efficiently building multiple serializers with derived types specified dynamically
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