Null Gulp pipe function

Question

I'm attempting to pass through a "dummy" pipe function that simply passes through to the next pipe instance based on a conditional. I cannot use gulp-if as I need sourcemaps support, so my current task looks like this:

module.exports = (gulp, $, browserSync, reload, merge, paths, files) ->
  ->

    nullFunc = ->
      transform = (file, cb) ->
        cb()
        return
      return

    uglify = if @seq[0] is 'coffeeprod' then $.uglify() else nullFunc()

    gulp.src(files.app)
    .pipe($.changed(paths.scripts.build))
    .pipe($.sourcemaps.init())
    .pipe($.concat('app.coffee'))
    .pipe($.coffee(join: true).on('error', $.util.log))
    .pipe(uglify)
    .pipe($.sourcemaps.write('../sourcemaps'))
    .pipe($.size(
      showFiles: true
      title: 'Coffee:'))
    .pipe gulp.dest(paths.scripts.build)

What's not working is my nullFunc() as I'm getting a TypeError: Cannot read property 'on' of undefined error. Any tips on a dummy function or the best way to get around the uglify() function not firing would be greatly appreciated. Thanks!