Proving equivalent categories have binary products if one of them does Suppose $\mathscr{C},\mathscr{D}$ are equivalent categories. 
Then there exist functors $S∶ \mathscr{C}\rightarrow \mathscr{D}$ and $T∶\mathscr{D}→\mathscr{C}$ with the compositions defined $T\circ S∶ \mathscr{C}→\mathscr{C}$ and $S\circ T∶ \mathscr{D}\rightarrow\mathscr{D}$, and a pair of natural isomorphisms $\alpha∶T\circ S \rightarrow 1_\mathscr{C}$ and $\beta ∶S\circ T\rightarrow1_\mathscr{D}.$
We seek to prove that, if $\mathscr{C} $ has binary products, then $\mathscr{D} $ also does.
I'm not really sure how to even approach this one, there seems to be so much going on that it's almost dizzying. My gut instinct is that we could show $\mathscr{C}$ has limits (since aren't limits a way of formalizing the product?) and then ... somehow? ... show that $\mathscr{D}$ also has them, but I'm so unconfident in that even conceptually.
Any ideas or nudges?
 A: Let $X,Y$ be objects in $\mathscr D$ and
\begin{align}
&p:T(X)\times T(Y)\to T(X)&
&q:T(X)\times T(Y)\to T(Y)
\end{align}
be a product source in $\mathscr C$.
We claim that
\begin{align}
&S(p)\beta_X:S(T(X)\times T(Y))\to X&
&S(q)\beta_Y:S(T(X)\times T(Y))\to Y
\end{align}
is a product source in $\mathscr D$.
Let
\begin{align}
&u:Z\to X&
&v:Z\to Y
\end{align}
be a source in $\mathscr D$.
Then
\begin{align}
&T(u):T(Z)\to T(X)&
&T(v):T(Z)\to T(Y)
\end{align}
is a source in $\mathscr C$, hence there exists one and only one morphism $w:T(Z)\to T(X)\times T(Y)$ such that
\begin{align}
&T(u)=wp&
&T(v)=wq
\end{align}
Then
\begin{align}
\beta^{-1}_ZS(w)S(p)\beta_X&=\beta^{-1}_Z(S\circ T)(u)\beta_X&
\beta^{-1}_ZS(w)S(q)\beta_Y&=\beta^{-1}_Z(S\circ T)(v)\beta_Y\\
&=\beta^{-1}_Z\beta_Zu&
&=\beta^{-1}_Z\beta_Zv\\
&=u&
&=v
\end{align}
which proves the existence part.
It remains to prove uniqueness.
For this, first recall that we can assume $\alpha^{-1}_TT(\beta)=1_T$ and $\beta^{-1}_SS(\alpha)=1_S$ (see, for example, here).
Let $w':Z\to S(T(X)\times T(Y))$ such that
\begin{align}
&u=w'S(p)\beta_X&
&v=w'S(q)\beta_Y
\end{align}
Then
\begin{align}
T(u)&=\alpha_{T(Z)}^{-1}T(\beta_Z)T(w')\alpha_{T(X)\times T(Y)}p&
T(v)&=\alpha_{T(Z)}^{-1}T(\beta_Z)T(w')\alpha_{T(X)\times T(Y)}q\\
&=T(w')\alpha_{T(X)\times T(Y)}p&
&=T(w')\alpha_{T(X)\times T(Y)}q\\
\end{align}
from which $w=T(w')\alpha_{T(X)\times T(Y)}$.
Consequently,
\begin{align}
S(w)
&=\beta_Z\beta^{-1}_Z(S\circ T)(w')S(\alpha_{T(X)\times T(Y)})\\
&=\beta_Zw'\beta^{-1}_{S(T(X)\times T(Y))}S(\alpha_{T(X)\times T(Y)})\\
&=\beta_Zw'
\end{align}
thus concluding the proof.
