Construction of Comonad Suppose $C$ has finite products and an object $I$. The exercise is to show that $T=I\times -$ is a comonad. Here is my attempt and several questions:
We will use $\pi$ for first projections and $\pi'$ for second projections, the domains being clear from the constructions. 
$1).$ Define the counit $\epsilon_b:I\times b\to b$ to be the second projection. 
$2).$ For the multiplication $\mu_b:I\times b\to I\times (I\times b)$ we will use the diagonal arrow $\delta:I\to I\times I$ so that $\mu_b$ is the unique arrow that satisfies $\pi\circ \mu_b=\delta\circ \pi$ and $\pi'\circ \mu_b=\epsilon_b.$ Of course, $\ \delta $ is the unique arrow that satifies $\pi\circ \delta =\epsilon_I\circ \delta =1_I.$
Using $1)$ and $2)$ I have shown that $\epsilon$ and $\mu$ are natural, using freely the fact that $I\times (I\times b)= (I\times I)\times b,$ the equality of course being shorthand for $\cong$.  My first question is, when I actually insert these isomorphisms, the required diagrams do not seem to commute. Where is my error? 
Next, I showed that $T$ is a comonad by verifying that the required diagrams commute. But I had to make claims that I cannot prove. For example, to show that $\epsilon_{Tb}\circ \mu_b=1_{I\times b}$ I used the naturality square 
\begin{array}{&&} 
I\times I\times b & \stackrel{\epsilon_{Tb}}{\to}& I\times b \\ 
\downarrow I\times \pi=\pi?& & \downarrow \pi \\ 
I\times I & \stackrel{\epsilon_I}{\to} & I
\end{array}
with $I\times \pi=\pi.$ But because $(I\times I)\times b $ is not the same thing as $I\times (I\times b)$, I have not been able to prove this. Can someone point me in the right direction?
 A: More generally, suppose $(C, \otimes)$ is a monoidal category. If $m$ is a monoid in $C$, then $m \otimes (-)$ is a monad, with multiplication inherited from the multiplication of $m$. Dually, if $n$ is a comonoid in $C$, then $n \otimes (-)$ is a comonad. 
Now show that if $C$ is a category with finite products, then every object is canonically a comonoid with respect to $\times$, where the comultiplication is the diagonal map $\Delta : c \to c \times c$. 
A: For the first question, it's really easier just to let the products strictly associate to the right, and let the comultiplication be $\nu_b:=\langle \pi,id_{I\times b}\rangle$. The naturality follows from how $(id_I\times(id_I\times f))\circ\langle \pi, id_{I\times b}\rangle$ shakes out. It should be easy to see that it is the same as $\langle\pi,id_{I\times c}\rangle\circ(id_I\times f)$ for $f:b\to c$.
The co-unit condition $\epsilon_{Tb}\circ\nu_b=id$ holds for the comultiplication I've defined above because you're just taking the second projection of a $\langle a,b\rangle$ where $b=id_{I\times b}$. The other half of this condition holds because $T\epsilon_b=id_I\times \pi'$, the composite of which with the comultiplcation ends up being $\langle\pi,\pi'\rangle=id$.
Adding the diagonal to any of this just complicates things needlessly, but if you were to do it, remember that the isomorphisms $(I\times I)\times b\cong I\times(I\times b)$ is not just natural, but actually has a particular description. From left to right, it's $$\langle \pi_{I\times I}\circ\pi_{(I\times I)\times b},\langle\pi'_{I\times I}\circ\pi_{(I\times I)\times b},\pi'_{(I\times I)\times b}\rangle\rangle,$$ and similarly for the other direction. It sounded like you were treating them as kind of blunt, amorphous natural isomorphisms, but really they have quite a lot of structure that you can unpack to understand why certain diagrams should commute.
