[proofplan] Exactness of sheaves means equality of the image sheaf and kernel sheaf at every position. Stalks commute with kernels, and the image sheaf is defined by sheafifying the presheaf image, whose stalk is the algebraic image on stalks. Therefore equality of image and kernel sheaves is equivalent to equality of their stalks at every point. [/proofplan] [step:Exactness of sheaves implies exactness on stalks] Suppose the sequence is exact at $\mathcal{F}^n$, so $\operatorname{im}(\mathcal{F}^{n-1}\to\mathcal{F}^n)=\ker(\mathcal{F}^n\to\mathcal{F}^{n+1})$ as sheaves. Taking stalks at $p$ gives \begin{align*} (\operatorname{im} d^{n-1})_p=(\ker d^n)_p. \end{align*} Stalks commute with kernels, so $(\ker d^n)_p=\ker(d^n_p)$. The stalk of the sheaf image is the image of the stalk map, so $(\operatorname{im}d^{n-1})_p=\operatorname{im}(d^{n-1}_p)$. Hence \begin{align*} \operatorname{im}(d^{n-1}_p)=\ker(d^n_p), \end{align*} which is exactness at $\mathcal{F}^n_p$. [/step] [step:Exactness on stalks identifies image and kernel sheaves] Conversely, assume the stalk sequence is exact for every $p$. At each position $n$, the composite $d^n\circ d^{n-1}$ has zero stalk at every point, hence is the zero sheaf morphism. Thus the image sheaf is contained in the kernel sheaf. For the reverse inclusion, exactness on stalks gives \begin{align*} (\operatorname{im}d^{n-1})_p=\operatorname{im}(d^{n-1}_p)=\ker(d^n_p)=(\ker d^n)_p \end{align*} for every $p\in X$. [/step] [step:Use the stalk-wise isomorphism criterion] The inclusion of sheaves $\operatorname{im}d^{n-1}\hookrightarrow\ker d^n$ induces an isomorphism on every stalk by the equality just proved. By the stalk-wise isomorphism criterion, this inclusion is an isomorphism of sheaves. Therefore $\operatorname{im}d^{n-1}=\ker d^n$ as sheaves at every position $n$, which is exactness of the original sequence. [/step]