Let a set $$A = A_1 \cup A_2 \cup \ldots \cup A_k$$, where $$A_i \cap A_j = \phi$$ for $$i \neq j$$; $$1 \leq i, j \leq k$$. Define the relation $$R$$ from $$A$$ to $$A$$ by $$R = \{(x,y) : y \in A_i$$ if and only if $$x \in A_i, 1 \leq i \leq k\}$$. Then, $$R$$ is:

Given: $$A = A_1 \cup A_2 \cup \ldots \cup A_k$$ with $$A_i \cap A_j = \phi$$ for $$i \neq j$$.

The relation $$R = \{(x, y) : y \in A_i \text{ if and only if } x \in A_i,\; 1 \leq i \leq k\}$$.

This means $$(x, y) \in R$$ if and only if $$x$$ and $$y$$ belong to the same partition $$A_i$$.

Reflexive: For any $$x \in A$$, there exists $$A_i$$ such that $$x \in A_i$$. Since $$x \in A_i \Leftrightarrow x \in A_i$$, we have $$(x, x) \in R$$. $$\checkmark$$

Symmetric: If $$(x, y) \in R$$, then $$x$$ and $$y$$ are in the same $$A_i$$. So $$y$$ and $$x$$ are also in the same $$A_i$$, giving $$(y, x) \in R$$. $$\checkmark$$

Transitive: If $$(x, y) \in R$$ and $$(y, z) \in R$$, then $$x, y$$ are in the same $$A_i$$ and $$y, z$$ are in the same $$A_j$$. Since $$y \in A_i$$ and $$y \in A_j$$, and the sets are disjoint, $$i = j$$. So $$x, z \in A_i$$, giving $$(x, z) \in R$$. $$\checkmark$$

Since $$R$$ is reflexive, symmetric, and transitive, it is an equivalence relation.

The answer is Option D: an equivalence relation.