# A question in the proof of equivalence of Cech cohomology and Sheaf cohomology

While proving the equivalence of Sheaf cohomology (defined by using injective resolutions), Tennison (in his book 'Sheaf Theory' on page 146) says :

If we let $S$ be a presheaf such that the sequence $0 \to R' \to R \to S \to 0$ is exact in Presh/X, we see that $S$ has as sheafification the zero sheaf and there is an exact sequence < long exact sequence corresponding to the above short exxact sequence>.

Here $R' \to R$ is the sheafification map of the presheaf $R'$. I can understand that $S$ can be shosen to be the Presheaf cokernel of the sheafification map. My question is : But how do we know that this map is exact at $R'$ ? Edit : As pointed out in the comment, this need not be exact at $R'$. So the question is : How do we tweak the proof in the book so that it becomes valid ?

• Unless $R^\prime$ is separated, meaning that two sections over an open which agree on some covering of that open must be the same, the map from $R^\prime$ to its sheafification will not be injective. – Keenan Kidwell Mar 2 '16 at 13:09
• @KeenanKidwell Thanks. That is right, I have edited the question to ask : how do we tweak the proof in the book then ? – user90041 Mar 4 '16 at 6:32
• You could consider instead the short exact sequences $0\to \text{ker}(s)\to R^{\prime}\to \text{im}(s)\to 0$ and $0\to \text{im}(s)\to R\to\text{coker}(s)\to 0$ where $s: R^{\prime}\to R$ is the sheafification map, using that the sheafifications of $\text{ker}(s)$ and $\text{coker}(s)$ vanish. – Hanno Mar 4 '16 at 7:03