Is the arrow category of cartesian closed category cartesian closed? in a cartesian closed category $\mathcal{C}$. if we have $f: A\to X$ and $g: B\to Y$


*

*then because the functor $(\_)^A$ is continuous, we have $g':B^A\to Y^A$

*composing $\text{id}\times f: Y^X\times A\to Y^X\times X$ and the evaluation map $\epsilon: Y^X\times X\to Y$ we have a function $Y^X\times A\to Y$ then we have induced a function $f': Y^X\to Y^A$

*we have the pushout of the two function $f'$ and $g'$ denoted $(Y, B)^{(X, A)}$


it seems that in the arrow category $\text{Id}\downarrow \text{Id}$ of $\mathcal{C}$, the object of type $(Y, B)^{(X, A)}\to Y^A$ in the pushout diagram is the exponential object of the two object $f$ and $g$. but I cannot formalize it, is this true?
What if restirct $f$ and $g$ to be monomorphism and replace the arrow category with the category of monomorphism pairs? notice that in this situation the morphisms in arrow category is only determined by the lower one of the arrow pair
 A: The first part of the question is essentially answered here. Your guess is half right: the codomain part of the exponential object $(Y, B)^{(X, A)}$ in $[\mathbb{2}, \mathcal{C}]$ is indeed just $Y^X$, i.e. the exponential object of the codomain parts. However the domain is more complicated; it is defined by the following pullback diagram:
$\require{AMScd}$
\begin{CD}
\bullet @>>> Y^X \\
@VVV @VV{Y^f}V\\
B^A @>>{g^A}> Y^A 
\end{CD}
Incidentally, the top horizontal arrow is the morphism part of the exponential object. 
Notice that if $g : B \to Y$ is a monomorphism then $g^A : B^A \to Y^A$ is a monomorphism, so we deduce that the full subcategory of $[\mathbb{2}, \mathcal{C}]$ spanned by the monomorphisms is closed under exponential objects. Since it is also closed under (finite) products, it follows that it is a cartesian closed category.
A: The arrow category of a category $\mathcal{C}$ is (isomorphic to) the category $\mathcal{C}^{\mathbf{2}}$, where $\mathbf{2}$ is the category $\bullet \rightarrow \bullet$ with two objects and one nontrivial morphism. The fact that this category is cartesian closed follows from a more general fact: if $\mathcal{C}$ is cartesian closed and $\mathcal{D}$ is any small category, then the category $\mathcal{C}^{\mathcal{D}}$ of functors $\mathcal{D} \to \mathcal{C}$ and natural transformations between them is cartesian closed. Finite products are given pointwise, and the exponential $F^G$ of functors $F,G : \mathcal{D} \rightrightarrows \mathcal{C}$ is defined by the exponential adjunction in $\mathcal{C}$: the correspondence is given by
$$\begin{matrix} (F^G)(A) \to TA \\ \hline FA \to GA \times TA \end{matrix}$$
naturally in $A \in \operatorname{ob} \mathcal{D}$ and $T : \mathcal{D} \to \mathcal{C}$.
A good exercise (for you) would  be to check what this means in the case $\mathcal{D} = \mathbf{2}$.
