$X^Y$ together with a morphism $\bf{ apply}$ $:X^Y\times Y\to X$ is an exponential object of $X$ and $Y$ if for each $Z$ and each morphism $f:Z\times Y \to X$, there is a unique morphism $\lambda f:Z\to X^Y$ such that $\lambda f \times id_Y \circ \bf {apply}$ $ = f$.
It is obvious that for every $f$ there is a $\lambda f$. But two things are not obvious to me:
Is there necessarily for every morphism $\lambda g:Z\to X^Y$ a corresponding morphism $g:Z\times Y \to X$? Couldn't we have a category where there are these $\lambda f:Z\to X^Y$, but also additional extraneous morphisms $\lambda g:Z\to X^Y$ that have no corresponding $g$? Can't we always add such extraneous morphisms?
Assuming that the previous question is clarified, and there is indeed a bijection between these homsets, why does wikipedia call it a "isomorphism" of homsets? An isomorphism is only an isomorphism within a category, and I don't know what category they're talking about.
(i) : No : start from $h : Z\to X^Y$; then you have $h\times id_Y : Z\times Y\to X^Y\times Y$, and then you can compose with $\mathbf{apply}$ to get $g:= \mathbf{apply}\circ (h\times id_Y) : Z\times Y\to X$.
Then by uniqueness, $\lambda g = h$.
(ii) : A bijection between sets is an isomorphism in the category of sets $\mathbf{Set}$.
But here it's even lore than that, because the isomorphism $\hom(Z\times Y, X) \cong \hom(Z, X^Y)$ is natural in all three variables, so it is actually an isomorphism between two functors in the category of functors $C^{op}\times C^{op}\times C\to \mathbf{Set}$