Let \(R\) be a near-ring. A fuzzy set in \(R\) is a mapping \(\mu\colon R\to[0,1]\). If \(\alpha\in[0,1]\) the set \(R^\alpha_\mu=\{x\in R\mid\mu(x)\geq\alpha\}\) is called a level subset of \(R\). \(\mu\) is a fuzzy subnear-ring if \(\mu(x-y),\mu(xy)\geq\min\{\mu(x),\mu(y)\}\) \(\forall x,y\in R\). A fuzzy subnear-ring \(\mu\) is a fuzzy ideal if \(\mu(y+x-y)\geq\min\{\mu(x),\mu(y)\}\), \(\mu(xy)\geq\mu(y)\) and \(\mu((x+z)y-xy)\geq\mu(z)\) \(\forall x,y,z\in R\). One-sided fuzzy ideals are defined in a natural way. If \(\mu\) is a fuzzy left (right) ideal of \(R\), then the level set \(R^\alpha_\mu\) is a left (right) ideal of \(R\), and hence is called a level left (right) ideal. If \(R\) and \(S\) are near-rings and \(f\colon R\to S\) is a mapping, and \(\mu\) is a fuzzy set of \(R\), then we define \(f(\mu)\) by \(f(\mu)(y)=\sup\{\mu(x)\mid x\in f^{-1}(y)\}\) if \(f^{-1}(y)\neq\emptyset\) and \(f(\mu)(y)=0\) otherwise. \(\mu\) is \(f\)-invariant if \(f(x)=f(y)\Rightarrow x=y\). Let \(f\colon R\to S\) be a near-ring epimorphism. It is shown that if \(\mu\) is a fuzzy left (right) ideal of \(R\) then \(f(\mu)\) is a fuzzy right (left) ideal of \(S\). Moreover, there is a one-to-one correspondence between the \(f\)-invariant left (right) ideals of \(R\) and \(S\). Furthermore, if \(\mu\), \(\nu\) are fuzzy left (right) ideals of \(R\) and \(S\) respectively, and \(\text{Im}(\mu)=\{\alpha_0,\alpha_1,\dots,\alpha_n\}\) with \(\alpha_0>\alpha_1>\cdots>\alpha_n\) and \(\text{Im}(\nu)=\{\beta_0,\beta_1,\dots,\beta_m\}\) with \(\beta_0>\beta_1>\cdots>\beta_m\) then (i) \(\text{Im}(f(\mu))\subset\text{Im}(\mu)\) and the chain of level ideals is \[ f(R^{\alpha_0}_\mu)\subset f(R^{\alpha_1}_\mu)\subset\cdots\subset f(R^{\alpha_n}_\mu)=S. \] (ii) \(\text{Im}(f^{-1}(\nu))\subset\text{Im}(\nu)\) and the chain of level ideals is \[ f^{-1}(S^{\beta_0}_\nu)\subset f^{-1}(S^{\beta_1}_\nu)\subset\cdots\subset f^{-1}(S^{\beta_m}_\nu)=R. \] A notion of prime fuzzy ideal is introduced. It is shown that if \(\mu\) is a prime fuzzy ideal, then \(\text{Im}(\mu)=\{0,1\}\).