What does this mean??

Question

say if something is a consistent mapping, then
$$ \forall \lambda,\mu\in\Lambda,\ f_{\lambda}|_{A_{\lambda}\ \cap A_{\mu}}=f_{\mu}|_{A_{\lambda}\ \cap A_{\mu}} $$ I can't understand what $$f_{\lambda}|_{A_{\lambda}\ \cap A_{\mu}}=f_{\mu}|_{A_{\lambda}\ \cap A_{\mu}} $$ means(by that I mean I can't understand the symbols)
so doe it have a proper definition?

Answer 1

The $f|A$ means $f$ restricted to $A$, which is just the function $f$ but where we only take $x$ values from the subset $A$ of its domain.

Also, two functions with a common domain are equal iff for all $x$ in that domain, their values agree on $x$.

So applying this the formula means:

$$\forall x \in A_\lambda \cap A_\mu: f_\lambda(x) = f_\mu(x)$$

The context: we have $f_{\lambda}: A_\lambda \to Y$ (or whatever the codomain is) and $f_\mu: A_\mu \to Y$. Then for all $x$ in $A_\lambda \cap A_\mu$ we have two possible candidate values for a combination function of $f_\lambda$ and $f_\mu$: $x \in A_\lambda$, so $f_\lambda(x)$ is defined, and $x \in A_\mu$ so $f_\mu(x)$ is also defined. As a function can only have one value for a given $x$ we need to have that actually $f_\lambda(x) = f_\mu(x)$ for such a combination function to be definable. As this holds for all $x$ in this intersection, we get $f_\lambda|_{A_\lambda \cap A_\mu} = f_\mu|_{A_\lambda \cap A_\mu}$ as the condition, in a shorthand way. In that case we can define $f: \cup_{\lambda \in L} A_\lambda \to Y$ by : if $x \in A_\lambda$, $f(x) = f_\lambda(x)$, and by the above condition we have that this does not depend on the $\lambda$, because if $x \in A_\mu$ as well, for $\mu \in L$, it would be in $A_\lambda \cap A_\mu$ and there $f_\lambda$ and $f_\mu$ agree on values.