I am looking at lecture notes in mathematical finance. The prince $C$ of a European call option with strike $K$ and expiry time $T$ in a one period binomial model is derived via no arbitrage principle. The underlying asset price is $S$ at time $0$ and either $Su$ or $Sd$ at time $1$ and interest rate is $r$. Lecture notes says that the return function $r_{i}(j)$ at present time (time $0$ I assume) is the following:
$$r(1) = \left( \frac{Su - K}{1+r} - C\right)$$
(if asset price moves up), or
$$r(2) = \left( \frac{Sd - K}{1+r} - C\right)$$
if the asset moved down. Then the no arbitrage principle gives equation $p_{1}r(1) + p_{2}r(2) = 0$ from which $C$ is found, where $p_{1,2}$ are risk neutral probabilities.
I don't get how the return function is derived. My understanding would be that, one gets a loan of $(C+K)$ money to purchase the option with price $C$ and also have the ability to pay strike price $K$. But then the accrued interest at time $1$ on the loan is $(1+r)(C+K)$ which we need to pay back; so this equivalently should give the present value return as
$$\tilde{r}(1) = \left( \frac{Su - K}{1+r} - C - K\right), \qquad \tilde{r}(2) = \left( \frac{Sd - K}{1+r} - C - K\right) $$
So what's going on?
You only need a loan of $C$. The $K$ part you obtain by exercising the option. Hence why the value of the option at time $t=1$ is either $Su-K$ or $Sd-K$. Actually, it should be $(Su-K)_{+}$ or $(Sd-K)_{+}$, because in case the underlying is worth less than the strike, you simply will not exercise the option.
Even if you take into account you want to make a loan beforehand to buy the underlying later, you make a loan of $K/(1+r)$ for that and hence that is precisely what is accounted for in the formula already. Which makes your reasoning double count the loan.
At time $t=0$, you make a loan to buy the option, your account looks like this:
$$\begin{array} {r|r} \text{Active} & \text{Passive} \\ \hline \text{Option } & \text{Loan } C + K/(1+r)\\ \text{Current account } K/(1+r) & \end{array}$$
At time $t=1-\epsilon$, just before exercising your option:
$$\begin{array} {r|r} \text{Active} & \text{Passive} \\ \hline \text{Option} & \text{Loan } C(1+r) + K\\ \text{Current account } K & \end{array}$$
At time $t=1$ you exercise the option, i.e. buy the underlying for K and sell it for $X$:
$$\begin{array} {r|r} \text{Active} & \text{Passive} \\ \hline & \text{Loan } C(1+r) + K\\ \text{Current account } X & \end{array}$$
At time $t=1+\epsilon$ you settle the loan, i.e. using $X$ to pay for it:
$$\begin{array} {r|r} \text{Active} & \text{Passive} \\ \hline & \text{Loan } C(1+r) + K - X = 0\\ \text{Current account } 0 & \end{array}$$
So, if you want to settle everything $X-C(1+r)-K=0$.