I'm reading Rotman - An Introduction to Homological Algebra: a great book! However I'm having some troubles understanding example 5.15 of page 225.
Let $X$ be a topological space and $(U_i)_{i\in I}$ a family of open subsets, and write $U_{ij} = U_i \cap U_j$ and $U = \bigcup_{i\in I} U_i$. It's easy to prove two facts (this is clear for me):
If $f,g : U\to \mathbb{R}$ are continuous functions such that $f_{\mid U_i} = g_{\mid U_i} \forall i\in I$, then $f=g$.
If we have a family of real valued functions such that ${f_i}_{\mid U_{ij}}={f_j}_{\mid U_{ij}} \forall i,j \in I$, then there exists a unique $f:U\to \mathbb{R}$ with ${f}_{\mid U_{i}} = f_i \forall i \in I$.
Now define $$\Gamma (V) = \left\{ \text{Continuous } \; f:V\to \mathbb{R} \right\}$$ and let $\Gamma(V)\to\Gamma(W)$ be the restriction map whenever $W\subset V$. Then properties 1 and 2 say that $\Gamma(U)$ is the equalizer of the family of maps $\Gamma(U_i)\to\Gamma(U_{ij})$.
Could you help me to see this last claim?
EDIT: My first trouble is that the definition of an equalizer deals with a family of maps $f_i: A \to B$. But here the above family of maps is not of this kind: the domains and codomains are not the same for every map.
$\Gamma(U)$ is the equalizer of $p \colon \prod \Gamma(U_i) \to \prod \Gamma(U_{i} \cap U_j)$ and $q \colon \prod \Gamma(U_i) \to \prod \Gamma(U_i \cap U_j)$, where $p$ maps $x_i$ in $x_{i |U_i \cap U_j}$ and $q$ maps $x_i$ into $x_{j|U_i \cap U_j}$. Can you verify it? To do it, just remember that $p$ (or $q$) is a product map, so they act component-wise.
I assumed that by $U_{ij}$ you mean the intersection of $U_i$ and $U_j$.
Just one more comment: this is true for any sheaf $\Gamma$ on a topological space. It might be true for sheaves on a site, but I don't exactly recall, please correct me if I'm wrong.