Let A be the set of all functions $$f: \mathbb{Z} \to \mathbb{Z}$$ and R be a relation on A such that $$R = \{(f, g) : f(0) = g(1) \text{ and } f(1) = g(0)\}$$. Then R is:

We are given the relation $$R = \{(f, g) : f(0) = g(1) \text{ and } f(1) = g(0)\}$$ on the set of all functions $$f: \mathbb{Z} \to \mathbb{Z}$$.

Reflexive: For $$(f, f) \in R$$, we need $$f(0) = f(1)$$ and $$f(1) = f(0)$$. Both conditions reduce to $$f(0) = f(1)$$. This is not true for all functions (e.g., $$f(x) = x$$ gives $$f(0) = 0 \neq 1 = f(1)$$). So R is not reflexive.

Symmetric: If $$(f, g) \in R$$, then $$f(0) = g(1)$$ and $$f(1) = g(0)$$. For $$(g, f) \in R$$, we need $$g(0) = f(1)$$ and $$g(1) = f(0)$$. The first condition follows from $$f(1) = g(0)$$, and the second from $$f(0) = g(1)$$. So R is symmetric.

Transitive: Suppose $$(f, g) \in R$$ and $$(g, h) \in R$$. Then $$f(0) = g(1)$$, $$f(1) = g(0)$$, $$g(0) = h(1)$$, $$g(1) = h(0)$$. For $$(f, h) \in R$$, we need $$f(0) = h(1)$$ and $$f(1) = h(0)$$. We have $$f(0) = g(1) = h(0)$$ and $$f(1) = g(0) = h(1)$$. So we need $$f(0) = h(1)$$ which means $$h(0) = h(1)$$, and $$f(1) = h(0)$$ which means $$h(1) = h(0)$$. This need not hold in general. For example, let $$f(0) = 1, f(1) = 0$$, $$g(0) = 0, g(1) = 1$$, $$h(0) = 1, h(1) = 0$$. Then $$(f,g) \in R$$ and $$(g,h) \in R$$, but $$f(0) = 1$$ and $$h(1) = 0$$, so $$f(0) \neq h(1)$$, meaning $$(f,h) \notin R$$. So R is not transitive.

Therefore R is symmetric but neither reflexive nor transitive.

Hence, the correct answer is Option B.