# Under what conditions does $Y^0 \cong 1$? (Category theory)

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.

Notation:
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?

• 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. Mar 25, 2019 at 16:47