Note: This is probably a duplicate, but I can't find the original question. If you find the original question, please link to it and then of course feel free to (vote to) close this question.

Question: If the category has an initial object $0$, a terminal object $1$, products, and if the indicated map (i.e. exponential) object exists, under what conditions does (for some objct $Y$) $$Y^0 \cong 1 \,?$$ Are there any conditions besides those above which are necessary for the isomorphism to hold?

I can show (see below) that $0 \times X \cong 0$ for all objects $X$ is sufficient, but is it also necessary? nLab says that when the category is distributive (which in particular implies that $0 \times X \cong 0$) then it does hold. But it would seem strange if this is a necessary condition, since Cartesian implies distributive. I.e. it seems like one should only have $Y^0 \cong 1 \implies 0 \times Y^0 \cong 0$, not vice versa.

Let $0 \times 1 \overset{\pi_{0 \times 1}^1}{\to} 0$ denote the "projection onto the first coordinate". Since $0 \times 1 \cong 0$ (since $A \times 1 \cong A$ for any $A$, we don't have to assume distributivity), and $\pi_{0 \times 1}^1$ is the isomorphism in one direction; the isomorphism in the other direction is $0 \overset{i_{0 \times X}}{\to} 0 \times X$, the unique map ("the $i$nitial map") $0 \to 0 \times X$. Let $Y \overset{t_{ Y^0} }{\to} 1$ denote the unique map ("the $t$erminal map") $Y^0 \to 1$. Denote the evaluation map for the map object by $e$, i.e. $0 \times Y^0 \overset{e}{\to} Y$. $\text{Id}$ denotes an identity morphism. Given $A \overset{f}{\to} B$ and $A \overset{g}{\to} C$, then $A \overset{\langle f , g \rangle}{\to} B \times C$ is the unique morphism (by the universal property of the product) such that $\pi_{B \times C}^1 \circ \langle f, g \rangle = f$ and $\pi_{B \times C}^2 \circ \langle f, g \rangle = g$. Then the notation $C_1 \times C_2 \overset{h_1 \times h_2}{\to } D_1 \times D_2$ denotes the morphism $h_1 \times h_2 = \langle h_1 \circ \pi_{C_1 \times C_2}^1 , h_2 \circ \pi_{C_1 \times C_2}^2 \rangle$, where we are assuming that $C_1 \overset{h_1}{\to} D_1$ and $C_2 \overset{h_2}{\to }D_2$.

Proof attempt:
Then the universal property of the map object gives us that, given the morphism $0 \times 1 \overset{i_Y \circ \pi_{0 \times 1}^1}{\to} Y$, there is a unique morphism $1 \overset{\phi}{\to} Y$ such that $e \circ (\text{Id}_0 \times \phi) = i_Y \circ \pi_{0 \times 1}^1$. Now clearly one has that $t_{Y^0} \circ \phi = \text{Id}_1$, so all that needs to be shown is that $\phi \circ t_{Y^0} = \text{Id}_{Y^0}$. It seems like the only way to go about doing that has to come down to using the universal property of $Y^0$ again, namely in this case that $\text{Id}_{Y^0}$ has to be the unique morphism such that $e \circ (\text{Id}_0 \times \text{Id}_{Y^0}) = e$. In other words, if we can show that $e \circ (\text{Id}_0 \times (\phi \circ t_{Y^0})) = e$, then it must follow that $\phi \circ t_{Y^0} = \text{Id}_{Y^0}$. Now one has that $$\begin{array}{rcl} e \circ (\text{Id}_0 \times (\phi \circ t_{Y^0})) &=& e \circ ((\text{Id}_0 \circ \text{Id}_0 ) \times (\phi \circ t_{Y^0}))\\ &= &e \circ (\text{Id}_0 \times \phi) \circ (\text{Id}_0 \times t_{Y^0})\\ &= &i_Y \circ \pi_{0 \times 1}^1 \circ (\text{Id}_0 \times t_{Y^0}) \\ & = & i_Y \circ \pi_{0 \times 1}^1 \circ \langle \pi_{0 \times Y^0}^1, t_{Y^0} \circ \pi_{0 \times Y^0}^2 \rangle \\ & = & i_Y \circ \pi_{0 \times Y^0}^1 \,. \end{array} $$

Now in general I don't see any reason at all why we should necessarily have that $ i_Y \circ \pi_{0 \times Y^0}^1 = e$. However, if $0 \times Y^0 \cong 0$, i.e. $0 \times Y^0$ were initial, then there would only be one morphism $0 \times Y^0 \to Y$, and since both $e$ and $i_Y \circ \pi_{0 \times Y^0}^1$ are morphisms $0 \times Y^0 \to Y$, it would follow from uniqueness that $e = i_Y \circ \pi_{0 \times Y^0}^1 \implies e \circ (\text{Id}_0 \times (\phi \circ t_{Y^0})) = e \implies \phi \circ t_{Y^0} = \text{Id}_{Y^0} \implies 1 \cong Y^0$.

Is there another way to solve this problem with the assumptions above that doesn't require assuming that $0 \times Y^0$ is an initial object? Or is $0 \times Y^0 \cong 0$ really a necessary condition?

  • 2
    $\begingroup$ One way to look at it : $Y^0 \simeq 1$ if and only if $\hom(-, Y^0)$ is a constant singleton, that is, if and only if $\hom(-\times 0, Y)$ is a singleton. So it is related to $X\times 0$. In fact, if for all $Y$, $Y^0\simeq 1$, then for all $X$, $X\times 0 = 0$. Now if you have a specific $Y$ for which $Y^0\simeq 1$, you can't use this of course. $\endgroup$ Mar 25, 2019 at 16:47


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