Let $S^1$ be the circle (with its usual topology), and let $\mathbb{Z}$ be the constant sheaf. Show that $H^1(S^1,\mathbb{Z})\cong\mathbb{Z}$.

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Rather than be given a proof, I'd like to know why my proof is wrong. (It's clearly wrong because I got the wrong final answer.)

We build an injective resolution of $\mathbb{Z}$; one possibility is

$$ 0\rightarrow\mathbb{Z}\rightarrow\mathbb{Q}\rightarrow\mathbb{Q}/\mathbb{Z}\rightarrow0\dotsb. $$

Indeed, this is exact, and $\mathbb{Q}$, $\mathbb{Q}/\mathbb{Z}$, and $0$ are injective $\mathbb{Z}$-modules because divisible. (We consider $\mathbb{Q}$, $\mathbb{Q}/\mathbb{Z}$, and $0$ as constant sheaves on $S^1$.)

The corresponding complex is

$$ \dotsb\rightarrow0\rightarrow \mathbb{Q}\overset{d^0}{\rightarrow} \mathbb{Q}/\mathbb{Z}\overset{d^1}{\rightarrow} 0\rightarrow \dotsb. $$

Applying the global section functor $\Gamma(S^1,\cdot)$, we get

$$ \dotsb\rightarrow\Gamma(S^1,0)\rightarrow \Gamma(S^1,\mathbb{Q})\overset{\Gamma(S^1,d^0)}{\rightarrow} \Gamma(S^1,\mathbb{Q}/\mathbb{Z})\overset{\Gamma(S^1,d^1)}{\rightarrow} \Gamma(S^1,0)\rightarrow \dotsb, $$

which just becomes our original sequence since we're dealing with constant sheaves.

Therefore, $H^1(S^1,\mathbb{Z})=\frac{\ker d^1}{im d^0}=\frac{\mathbb{Q}/\mathbb{Z}}{\mathbb{Q}/\mathbb{Z}}=0$.