## Appendix

##### A.1. Proof of Proposition 1. Pure strategy equilibria of the product improvement game

The text shows that, if the firms are restricted to invest either $ z^{M} $ or 0, then there is an equilibrium in which Firm 2 invests $ z^{M} $ and Firm 1 invests 0. In addition, if and only if $ r(z^{M}) \geq (\gamma_{2} - \gamma_{1})M $ , there is another equilibrium in which Firm 1 invests $ z^{M} $ and Firm 2 invests 0.

There are no other profitable deviations when firms' choice sets are unrestricted. Investment by Firm 1 cannot be profitable unless it can win the market from Firm 2, which requires $ z_{1} \geq z^{M} + (\gamma_{2} - \gamma_{1}) $ when Firm 2 invests $ z^{M} $ . The best investment for Firm 1 in this case maximizes $ \pi_{1} = (z_{1} - z^{M} - \gamma_{2} + \gamma_{1} + w)M - r(z_{1}) $ subject to this constraint. Convexity of $ r(z) $ implies that the constraint binds and Firm 1's maximum deviation profit is $ \pi_{1} = wM - r(z^{M} + \gamma_{2} - \gamma_{1}) < wM $ . Thus Firm 1 earns less profit by deviating from $ z_{1} = 0 $ when Firm 2 invests $ z^{M} $ , and given that Firm 1 chooses $ z_{1} = 0 $ , the profit-maximizing investment for Firm 2 is $ z_{2} = z^{M} $ . If $ z_{1} = z^{M} $ and $ z_{2} = 0 $ , Firm 1 has no incentive to deviate and earn $ \pi_{1} = wM $ by choosing $ z_{1} = 0 $ , and has no incentive to choose any other level of quality improvement. If it were profitable for Firm 2 to deviate, then Firm 2 would choose $ z^{M} $ and earn $ \pi_{2} = (\gamma_{2} - \gamma_{1})M - r(z^{M}) $ . Therefore, Firm 2 has no incentive to deviate if and only if $ r(z^{M}) \geq (\gamma_{2} - \gamma_{1})M $ .

Nor are there any other candidate equilibria when firms' choice sets are unrestricted. For example, if $ \gamma_{2} + z^{M} > \gamma_{2} + z_{2} \geq \gamma_{1} + z_{1} $ , then Firm 2 would deviate profitably by investing $ z^{M} $ instead of $ z_{2} $ . Alternatively, if $ \gamma_{2} + z_{2} > w $ and $ \gamma_{2} + z_{2} > \gamma_{1} + z_{1} > \gamma_{1} $ , then Firm 1 would deviate profitably by investing 0 instead of $ z_{1} $ . Other cases trigger similar deviations to $ z^{M} $ or 0, depending on whether the firm wins the market or not.

##### A.2. Proof of Proposition 2. Pure strategy equilibria of the technological tying game

The text identifies the equilibria of the technological tying game for various cases under the assumption that the firms' investment choices are restricted to $ z^{M} $ and 0. Alternatively, consider any non-negative investment levels $ (z_{1}, z_{2}) $ . It is straightforward to show that Firm 1's (Firm 2's) best response to any non-negative $ z_{2}(z_{1}) $ always belongs to $ \{0, z^{M}\} $ , i.e. any $ z_{1}(z_{2}) \neq \{0, z^{M}\} $ is strictly less profitable for Firm 1 (Firm 2) than $ z^{M} $ or 0. Consequently, there are no other profitable deviations and no other candidate equilibria when firms' choice sets are unrestricted.

Suppose, for example, that $ w \leq \gamma_{2} $ . Then $ w < \gamma_{1} + z^{M} - r(z^{M}) / M $ by A3. If Firm 1 were to invest $ z_{1} $ and foreclose Firm 2 with a technological tie, then Firm 1 would earn $ \pi_{T}(z_{1}) = (\gamma_{1} + z_{1}) M - r(z_{1}) $ . This profit function reaches a maximum at $ z^{M} $ . If Firm 1 did not foreclose Firm 2 with a technological tie, then Firm 1 would