Presheaf with same global sections as associated sheaf

Question

My question is very easy to state : if a presheaf has the same global sections as its associated sheaf, is it a sheaf ?

I imagine its false but havent found a counter example.

Answer 1

Consider the sheaf $ \mathcal C$ of continuous functions on $[0,1]$ and its subpresheaf $\mathcal C_b\subset \mathcal C$ of bounded continuous functions .
The sheaf associated to $\mathcal C_b$ is $\mathcal C$ (because every continuous function is locally bounded) but we have $\mathcal C_b\subsetneq \mathcal C$, since $\frac 1x\in \mathcal C(0,1)\setminus \mathcal C_b(0,1)$.
However the presheaf and the sheaf have the same global sections $\mathcal C_b[0,1]=\mathcal C[0,1]$, since a continuous function on the compact interval $[0,1]$ is necessarily bounded.