The title is fairly self explanatory. Let $\mathcal{A}$ be a small category , $X$ be an object of $\widehat{\mathcal{A}}$. Then the question is, are the categories, $\widehat{\mathcal{A}\downarrow X}$, and $\widehat{A}\downarrow X$ equivalent? I am 99% sure that this is true. Furthermore, if their is a reference for this, that would be helpful.

By $\widehat{\mathcal{A}}$ you must mean the category of presheaves on $\mathcal{A}$. I take it that $X$ denotes a presheaf, and that $\mathcal{A}/X$ is the comma category $y_{\mathcal{A}} \downarrow X$ where $y_{\mathcal{A}}$ is the Yoneda embedding. Another reference for this fact is Tom Leinster's book Higher Operads, Higher Categories, p. 394. I'm not sure who first discovered it, but it's ancient folklore. 


Actually, there is an abstractnonsense proof (although that doesn't make constructing the equivalence by hand any less of a good exercise). It goes through the equivalence of presheaves on a category A with discrete fibrations over A, so $\widehat{A} \simeq \mathrm{DFib}(A) \subset (\mathrm{Cat}\downarrow A)$. Note that DFib(A) is a full subcategory of $\mathrm{Cat}\downarrow A$: any map between discrete fibrations is a map of fibrations. Furthermore, a composite of discrete fibrations is a discrete fibration, and conversely any map between discrete fibrations is itself a discrete fibration. Thus, for any discrete fibration B → A, we have $\mathrm{DFib}(A)\downarrow B \simeq \mathrm{DFib}(B)$. Thus, if $B = A\downarrow X$ is the category of elements of a presheaf $X\colon A^{op} \to \mathrm{Set}$, then we have $(\widehat{A}\downarrow X) \simeq (\mathrm{DFib}(A)\downarrow B) \simeq \mathrm{DFib}(B) \simeq \widehat{B} = \widehat{A \downarrow X}$ See also this question. 


I'm guessing $\widehat{A}$ denotes the presheaf category? If so, see Johstone's "Sketches of an Elephant" page 8. 

