$x\in X$, then let $s(x) = m_x(r(f(x)))$. Then, $s$ is a section of $\shf{S}$ because whenever $x \le y \in X$, $$\begin{aligned}
\left(\shf{S}(x \le y)\right)s(x) &=& \left(\shf{S}(x \le y)\right) m_{x}(r(f(x)))\\
&=& \left(\shf{S}(x \le y) \circ m_{x}\right)(r(f(x)))\\
&=& \left(m_y \circ \shf{R}(f(x) \le f(y)) \right)(r(f(x)))\\
&=& m_y (r(f(y)))\\
&=&s(y)\end{aligned}$$ by the definition of a sheaf morphism.

Generalizing a bit further, it is also useful to be able to map the stalks of a dual sheaf into the stalks of a sheaf – providing a notion of a *hybrid morphism* from dual sheaves into sheaves. This plays an important role in understanding discretizations.

\[df:hybridmorphism\] Suppose that $\dshf{D}$ is a dual sheaf on a poset $Y$ and that $\shf{S}$ is a sheaf on $X$. A *hybrid morphism* $m: \dshf{D}\to \shf{S}$ along an order preserving map $f: X \to Y$ consists of a set of functions $m_x: \dshf{D}(f(x)) \to \shf{S}(x)$ for each $x \in X$ such that the following diagram commutes $$\xymatrix{
      \dshf{D}(f(y)) \ar[r]^{m_y} \ar[d]_{\dshf{D}(f(x) \le f(y))} &\shf{S}(y)\\
      \dshf{D}(f(x)) \ar[r]_{m_x} &\shf{S}(x) \ar[u]_{\shf{S}(x \le y)}\\
      }$$ for each $x \le y$. We usually call the functions $m_x$ the *components* of the hybrid morphism. A hybrid morphism is said to be *injective* (or *surjective*) if each component 