Product of skyscraper sheaf vs constant sheaf Let $X$ be a manifold. The inclusion $i_x:x \rightarrow X$ is continuous (unless I am going crazy). Now let $A$ be an abelian group and let $\mathcal{A}_x$ be the constant sheaf $A$ on $\{x\}$ and let $\mathcal{A}$ denote the sheaf $\prod_{x\in X} (i_x)_*(\cal{A}_x)$. Then $\mathcal{A}(U) = \prod_{x\in U} A$. Hopefully I haven't made any errors so far.
Now what I cannot figure out is this: if $\mathcal{F}$ denotes the constant sheaf $A$ on $X$, then there is a canonical map $\mathcal{F}\rightarrow \cal{A}$ given by $s \mapsto (s,s,...,s)$ locally. Now on the level of stalks, this is just the identity $A \rightarrow A$ and is hence an isomorphism. However, the two sheaves are clearly not isomorphic; indeed if $U$ is connected, open and contains more than two points, then $\mathcal{F}(U) = A$ but $\mathcal{A}(U) = \prod_{x\in U} A$, so I've clearly made a mistake somewhere. Any help would be appreciated!
 A: Your issue is the claim that the stalk of $\mathcal{A}$ at any point is simply $A.$ Let's look at an example: take $X = \Bbb{R}$ and $A = \Bbb{Z}$ for concreteness, and let's look at the stalk of $\mathcal{A}$ at $0.$
For any sheaf $\mathcal{F}$ on any space $X,$ we have an explicit description of the stalk of $\mathcal{F}$ at a point $x$:
$$
\mathcal{F}_x \cong \{\sigma\in \mathcal{F}(U)\mid U\ni x\}/\sim,
$$
where $\sigma\in \mathcal{F}(U)$ and $\tau\in\mathcal{F}(V)$ are equivalent if there exists some open $W\ni x$ such that $\left.\sigma\right|_W = \left.\tau\right|_W.$
So, for our particular $\mathcal{A},$ an element of $\mathcal{A}_0$ is represented by some $(\sigma_x)_{x\in U}$ with $U\ni 0$, where each $\sigma_x\in\Bbb{Z}.$ There is a morphism of groups
\begin{align*}
e : \mathcal{A}_0&\to \Bbb{Z}\\
[(\sigma_x)]&\mapsto\sigma_0.
\end{align*}
The kernel of this map consists of $[(\sigma_x)]$ such that $\sigma_0 = 0.$ However, I claim that $e$ is not injective. In particular, if we let $U = (-\epsilon,\epsilon)$ for some $\epsilon > 0$ and set
$$
\tau_x := \begin{cases}
0,&x = 0,\\
1,&x\neq 0.
\end{cases}
$$
If $\tau = (\tau_x)_{x\in U},$ then $[\tau]\in\ker e,$ but $\tau\not\sim 0$: we have $\left.\tau\right|_V = (\tau_x)_{x\in V}\neq (0)_{x\in V}$ for any open $V\ni 0,$ since any open neighborhood of $0$ contains nonzero real numbers.

In fact, notice that
\begin{align*}
\mathcal{A}(U)&\to\{f : U\to A\}\\
s = (s_x)_{x\in U}&\mapsto\left.\begin{cases}f_s : &U&\to A\\
&x&\mapsto s_x\end{cases}\right\}
\end{align*}
is an isomorphism for any $U.$ That is, your sheaf is the sheaf of functions into $A.$ One common interpretation of the elements of the stalk of a sheaf of functions is as "germs of those functions." With this interpretation, it might be clearer that the stalk of $\mathcal{A}$ at a point $x$ is not simply $A,$ since a germ of a function $f$ at a point $x$ is not determined only by the value of $f$ at $x,$ but also by "infinitesimal data" of $f$ near $x.$
