$\pi:\Bbb P^1_k\to \text{Spec}(k)$ then is $\Bbb L\pi^*\tilde{k}\cong \mathcal{O}_{\Bbb P^1_k}[0]$?

Question

Let $\pi:\Bbb P^1_k\to \text{Spec}(k)$. Am I correct that $\Bbb{L}\pi^*\tilde{k}$ is just $\mathcal{O}_{\Bbb{P}^1_k}[0]\in D_{\text{qc}}(\Bbb P^1_k)$? I saw $\Bbb{L}\pi^*\tilde{k}$ being talked about in some lecture notes, but I was confused why this wasn't identified with $\mathcal{O}_{\Bbb P^1_k}[0]$.

Here is my reasoning:

1. I think $\tilde{k}$ is just $\mathcal{O}_{\text{Spec}(k)}$,
2. I think that for any scheme map $\pi:X\to Y$ it is true that $\pi^*\mathcal{O}_Y\cong \mathcal{O}_X$ (as $\mathcal{O}_X$-modules),
3. I think that $\tilde{k}$ is free so projective, so $0\to \tilde{k}\to \tilde{k}\to 0$ is a projective resolution with augmentation map of $\tilde{k}$,
4. I think this means $\Bbb L\pi^*\tilde{k}$ is just $\pi^*(\tilde{k}[0])= [\dots \to 0\to \pi^*\mathcal{O}_{\text{Spec}(k)}\to 0\to\dots] = \mathcal{O}_{\Bbb P^1_k}[0]$.

Is anything wrong. Thanks! I am newbie to algebraic geometry and only self study.