Why is the class $[i^*i_*\mathcal{F}]$ not equal to $i^*i_*[\mathcal{F}]$ in K-theory?

Question

Let $i:X\to Y$ be a closed embedding of smooth varieties and $\mathcal{F}$ be a coherent sheaf of $\mathcal{O}_X$-modules. Then why is $i^*i_*[\mathcal{F}]\neq[\mathcal{F}]=[i^*i_*\mathcal{F}]$ in $K_0(X)$?

EDIT: the previous version of this question asked why $i^*i_*\mathcal{F}\neq\mathcal{F}$ and included a proof that $i^*i_*\mathcal{F}=\mathcal{F}$ until Roland helped me understand the question was wrongly formulated.

Answer 1

Here is a counter-example, I hope it will show why the equality does not hold.

Let $X=0$ and $Y=\mathbb{A}^1$ as in your post. Let $\mathcal{F}=\mathcal{O}_X=k$. Then $i_*\mathcal{F}$ is not a vector bundle, so we need to find a resolution by vector bundles. Here is one : $$ 0\rightarrow\mathcal{O}_Y\overset{\times t}\rightarrow\mathcal{O}_Y\rightarrow i_*k\rightarrow 0$$

So $[i_* k]=i_*[k]=[\mathcal{O}_Y]-[\mathcal{O}_Y]=0$. Hence $i^*i_*[k]=0$.