Given a category with exponentials, $\lambda f \circ g = \lambda (f \circ (g \times id))$ I'm trying to prove the next lemma, and I can't seem to find a solution, even though it looks pretty easy. The lemma is, given a category with exponentials, $\lambda f \circ g = \lambda (f \circ (g \times id))$.
The definition I've got of the exponential object between $A$ and $B$ is an object $B^A$ together with an arrow $ev:B^A \times A \rightarrow B$, and for every object $C$ and arrow $f:C \times A \rightarrow B$ there is an unique arrow $\lambda f:C \rightarrow B^A$ such that $ev \circ (\lambda f \times id_A) = f$.
 A: Let me try fixing context and notation first. So ${\mathscr C}$ is a Cartesian closed category, and if $f: B\times X\to Y$ is a morphism in ${\mathscr C}$, we denote $\lambda_X f: B\to Y^X$ the adjoint/$\lambda$-closure of $f$ with respect to $X$. Then, given $g: A\to B$, you're asking for a proof of 
$$(\lambda_X f)\circ g = \lambda_X (A\times X\xrightarrow{g\times\text{id}_X} B\times X\xrightarrow{f} Y),$$ right? 
You can prove this by checking that the left hand side $(\lambda_X f)\circ g$ satisfies the characterizing property of the right hand side: Namely, by your definition of $\lambda$, the morphism $$h := \lambda_X (A\times X\xrightarrow{g\times\text{id}_X} B\times X\xrightarrow{f} Y): A\to Y^ X$$ is uniquely determined by the equation $$A\times X\xrightarrow{h\times\text{id}_X} Y^X\times X\xrightarrow{\text{ev}} Y\ =\ A\times X\xrightarrow{g\times\text{id}_X} B\times X\xrightarrow{f} Y$$ and by writing things out carefully you can check that $(\lambda_X f)\circ g$ has this property.
