The objectwise image of a sheaf morphism need not be a sheaf

Let $U_1=S^1\setminus \{(-1,0)\}$ and $U_2=S^1\setminus \{(1,0)\}$. These open arcs cover $S^1$. On each $U_i$ choose a continuous argument ${\theta }_i:U_i\to \mathbb{R}$ with $e^{2\pi i{\theta }_i(z)}=z$. Therefore the identity map ${\mathrm{id}}_{S^1}$ restricts to sections in the objectwise image presheaf on both $U_1$ and $U_2$.

On the overlap $U_1\cap U_2$, both local sections are equal to the same target section ${\mathrm{id}}_{S^1}{\mathrm{\mid }}_{U_1\cap U_2}$, so they are compatible.

Suppose ${\mathrm{id}}_{S^1}$ lay in the global objectwise image. Then there would be a continuous $f:S^1\to \mathbb{R}$ with $e^{2\pi if(z)}=z$ for every $z\in S^1$. Writing $z=e^{2\pi it}$ with $t\in [0,1]$, the function $g(t):=f(e^{2\pi it})-t$ is continuous and integer valued, hence constant. So $f(e^{2\pi it})=t+n$ for some fixed integer $n$. Evaluating at $t=0$ and $t=1$ gives two values of $f$ at the same point $1\in S^1$, namely $n$ and $n+1$, a contradiction. Thus ${\mathrm{id}}_{S^1}$ is not in the global objectwise image.

Steps 1.1 to 3.1 give compatible local sections in the image presheaf that do not glue globally, so the objectwise image is not a sheaf and hence not a subsheaf by [F1]. By [L1], its sheafification is the correct image sheaf.