Make two files instead of three

Hey!

I effectively want to split out part of a file, instead of splitting out the common parts of two files.

Currently I have bundle.js, which contains everything used on the site. There is some checkout specific logic there that I only want on the checkout, so I want to split that stuff out into bundle-checkout.js, but then run both bundle.js and bundle-checkout.js in the checkout.

With factor-bundle by default, I would be running these files on the main site:
- bundle-common.js
- bundle.js

And these on the checkout:
- bundle-common.js
- bundle.js
- bundle-checkout.js

It's pointless for me to have both bundle-common.js and bundle.js!

The best solution I've found so far is to add `require('./main')` to the top of the input `bundle-checkout.js`, which would mean that `bundle.js` is effectively an empty file and I can include `bundle-common.js` by itself, but then that generates a junk file.

---

lib.js:

``` js
module.exports = 'lib';
```

main.js:

``` js
var lib = require('./lib');
console.log(lib);
```

checkout.js:

``` js
require('./main');
console.log('special checkout logic');
```

And then the command:

```
browserify main.js checkout.js -p [ factor-bundle -o junk.js -o bundle-checkout.js ] -o bundle.js
```

Is there any way to do this without the `require('./main')` and the useless `junk.js` file?

Thanks, and sorry for the long question! I'll turn this into an article once it's been figured out so that nobody asks it again :)

Comienza con el comando factor-bundle mostrado en el issue y reprodúcelo usando lib.js, main.js y checkout.js. Traza cómo el comando gestiona los bundles compartidos y de entrada; se considera completado cuando genera los bundles solicitados del sitio y del checkout sin requerir ./main ni producir junk.js.

Escrito por el modelo de indexación a partir del texto del issue.