|
|
|
|
|
by unalone
6143 days ago
|
|
They partnered with AT&T because they needed a company that would let it add things like visual voicemail to its existing system, and rolling that out with a single company would be easier. The argument against Verizon is that Verizon insists on adding branding to every Verizon phone, which Apple is against. Neither Verizon nor Apple is stupid. If network choice really starts hurting the iPhone, which it isn't because iPhone sales are insane, then Apple will bend slightly to accommodate Verizon. If AT&T continues gaining Verizon users on virtue of the iPhone itself, then Verizon will look for a deal with Apple. The point of the ancestor post is that if your strategy to compete is "we're shittier but use a different network", then you aren't focused on the product, you're focused on making money. John Gruber is writing about product, not about market competition. |
|
The exclusivity was about a lot more than visual voice mail. It had to do with sales channel and profits from ongoing contracts as well, which were probably far larger issues and ones Verizon has absolutely no reason to cave on. And my point was that it might be in both Verizon and Apple's best interest to not work together.
The network is a huge part of the product when it comes to phones. A phone that is constantly dropping calls, one of the iPhone's biggest complaints, or out of fast data range is an inferior product through no fault of its own. In the mobile industry by being on a better network you have a better product.
Nonetheless, the best and worst thing about capitalism is that inferior products win all the time when the salient point of competition isn't product "quality" as you're meaning it here. We in the tech industry love the idea of a Google or an Apple making a better mousetrap and slaying the entrenched competition, but the reality is that for every one story like that, there are 20 of a better product that died due to vendor lock-in, marketing, or some other form of differentiation. Apple's been on both sides of that equation.