A functor sends $X$ to $R^{(X)}$ what does this notation mean? Is this functor faithful and full?
Functor from $\textbf{set}$ to $R\textbf{-module}$
1
$\begingroup$
category-theory
-
0I prefer to denote this $R\langle X \rangle$, because it's a lot like the span construction from linear algebra, which is sometimes denoted $\langle X \rangle$. – 2017-01-31
2 Answers
2
Usually, $R^{(X)}$ denotes the free $R$-module on the set $X$, that is, the set of “formal sums” $$ \sum_{x\in X}r_xx $$ where $\{x\in X:r_x\ne0\}$ is finite, with the obvious operations making it into an $R$-module. The elements of $X$ are those where just one coefficient is $1$ and the others are zero.
How do we make the assignment $X\mapsto R^{(X)}$ into a functor? Use the universal property of free modules: if $Y$ is any set, then for each map $f\colon X\to Y$ there is a unique $R$-module homomorphism $\hat{f}\colon R^{(X)}\to R^{(Y)}$ such that $\hat{f}(x)=f(x)$, for every $x\in X$: $$ \hat{f}\Bigl(\,\sum_{x\in X}r_xx\Bigr)=\sum_{x\in X}r_xf(x) $$ where the sum in the right hand side is well defined because at most a finite number of terms are nonzero.
Clearly the identity is mapped to the identity and this assignment preserves composition of morphisms.
The functor so obtained is certainly faithful. In order to see whether it is full, look at the case when $X$ is a singleton.
1
$R^{(X)}$ denotes the free R-module over X.
The functor is faithful since different maps on bases of free modules induce different module homomorphisms.
It is not full since there exist module homomorphisms between free modules that are not induced by maps on their bases (e.g. the zero homomorphism).