Comparative statics question with an application In the state of Mexas, two politicians (Mr. BO, or "Politician 1" and Mr. TC, or "Politician 2") are competing intensely for a senate seat. The two politicians spend on advertising to increase the number of supporters. A political consultant finds that the optimal advertising expenditure of Mr. BO, $S_{1}$, depends on the spending $S_{2}$ by Mr. TC and a "likability" parameter $\alpha$ that influences the popularity of Mr. BO in Mexas: $$\text{Equation 1:  }S_{1} = f\left ( S_{2}, \alpha \right ),$$
where $f:\mathbb{R}_{+}^{2}\rightarrow \mathbb{R}$ is twice continuously differentiable, $0&lt;\frac{\partial f\left ( S_{2}, \alpha \right )}{\partial S}&lt;1$ and $0&lt;\frac{\partial f\left ( S_{2}, \alpha \right )}{\partial \alpha}&lt;1$ for all $S_{2} \geq 0$ and all $\alpha \geq 0$. 

The optimal expenditure of politician 2 depends on the spending $S_{1}$ of politician 1 and a "redness" parameter $\beta$, which influences how many people would stick to Mr. TC: $$\text{Equation 2:  }S_{2} = g\left ( S_{1}, \beta \right ),$$
where $g:\mathbb{R}_{+}^{2}\rightarrow \mathbb{R}$ is twice continuously differentiable, $0&lt;\frac{\partial g\left ( S_{1}, \alpha \right )}{\partial S}&lt;1$ and $0&lt;\frac{\partial g\left ( S_{1}, \alpha \right )}{\partial \alpha}&lt;1$ for all $S_{1} \geq 0$ and all $\beta \geq 0$. 

The equilibrium values of $S_{1}$ and $S_{2}$ are given by the solution of the simultaneous equations (1) and (2). Suppose that there is a unique solution $S_{1}^{*}&gt;0$ and $S_{2}^{*}&gt;0$. 

Does an increase in $\alpha$(holding $\beta$ constant) necessarily increase or necessarily decrease $S_{1}^{*}$? Explain. 

My attempt: Here I have used Implicit Function Theorem to answer the question, as we are essentially looking for the comparative statics of $\frac{\mathrm{d} S_{1}^{*} }{\mathrm{d} \alpha }$. 

Since from equation 2, $S_{2}^{*} = g\left ( S_{1}^{*}, \beta \right )$, I substituted $S_{2}^{*}$ in equation to obtain $$S_{1}^{*} = f\left ( g\left ( S^{*}_{1}, \beta \right ), \alpha \right ).$$ Then since $$S_{1}^{*} - f\left ( g\left ( S^{*}_{1}, \beta \right ), \alpha \right ) = 0 \equiv F,$$
I can apply the Implicit Function Theorem (IFT) to derive $\frac{\mathrm{d} S_{1}^{*} }{\mathrm{d} \alpha }$.

So $$\frac{\mathrm{d} S_{1}^{*} }{\mathrm{d} \alpha } = - \frac{\frac{\partial F}{\partial \alpha}}{\frac{\partial F}{\partial S_{1}}} = - \frac{\frac{\partial f\left (  \right )}{\partial \alpha}}{1 - \frac{\partial f \left (  \right )}{\partial g} \frac{\partial g \left (  \right )}{\partial S_{1}^{*}}} \frac{&gt;}{&lt;} 0.$$

Since we do not know the sign of the derivative $\frac{\partial f \left (  \right )}{\partial g}$, the effect of $\alpha$ on $S^{*}_{1}$ is ambiguous. 

Does my answer make sense?
You can obtain a definite answer for $$\text{sign}\left\{\frac{\mathrm{d} S_{1}^{*} }{\mathrm{d} \alpha }\right\}$$ given the assumptions.

From

$$S_{1}^{*} - f\left ( g\left ( S^{*}_{1}, \beta \right ), \alpha \right ) = 0 \equiv F$$

and the Implicit Function Theorem

$$\frac{\mathrm{d} S_{1}^{*} }{\mathrm{d} \alpha } = - \frac{\partial F/\partial \alpha}{\partial F/\partial S^*_{1}} $$

we have  

$$\frac{\partial F}{\partial \alpha} = -\frac{\partial f}{\partial \alpha}$$

and 

$$\frac{\partial F}{\partial S^*_{1}} = 1-\frac{\partial f}{\partial S_2}\cdot \frac{\partial g}{\partial S^*_{1}}$$

For these expressions we know not only the signs but also the magnitudes. The result follows.