Just before the truth-arrows in a topos subsection of Goldblatt's "Topoi: A Categorial Analysis of logic," descriptions of the truth functions $\Rightarrow$ and $\smallsmile$ are given in $\mathbf{Set}$ in terms of arrows. This is to motivate their abstraction to any given topos.

I'm having trouble understanding these descriptions thoroughly, so I would like some clarification, please :)

Definition 1: The disjunction truth-function $\smallsmile :2\times 2\to 2$ (for $2=\{0, 1\}$) is given by $1\smallsmile 1=0\smallsmile1=1\smallsmile0=1$ and $0\smallsmile0=0$, whereas the implication truth-function $\Rightarrow: 2\times 2\to 2$ is given by $1\Rightarrow0=0$ and $(0\Rightarrow1)=(0\Rightarrow0)=(1\Rightarrow1)=1$.

Definition 2: $\smallfrown$ can be taken as the characteristic function of the product map $\langle\operatorname{true}, \operatorname{true}\rangle$.


Goldblatt states that $\Rightarrow$ is the characteristic function of enter image description here$=\{\langle 0, 0\rangle,\langle 0, 1\rangle, \langle 1, 1\rangle\}$ with

enter image description here

a pullback square, which he then uses to write enter image description here as the equaliser of $\smallfrown:2\times 2\to 2$ and $pr_1$, where $pr_1(\langle x,y\rangle)=x$. I don't see how he does this.


We take $\smallsmile$ as $\chi_D$ for $D=A\cup B$ for $A=\{\langle1, 1\rangle, \langle1, 0\rangle\}$ and $B=\{\langle1, 1\rangle,\langle0, 1\rangle\}$; identify $A$ with the product map $\langle\operatorname{true}_2, 1_2\rangle$, $B$ with $\langle1_2, \operatorname{true}_2\rangle$; form the coproduct map $f=[\langle\operatorname{true}_2, 1_2\rangle, \langle1_2, \operatorname{true}_2\rangle]$; identify $\operatorname{im} f=D$; epic-monic factor $f$ through $D$; then claim that all this specifies $D$ - and hence $\small\smile$ - uniquely up to isomorphism.

So yeah, I can't see the wood for the trees basically. Please help :)

  • $\begingroup$ Okay, things are starting to make sense . . . $\endgroup$ – Shaun Jun 28 '14 at 18:18
  • $\begingroup$ Ah, I get it. It's easy! I'm overthinking it. I'll type up my answer tomorrow :) $\endgroup$ – Shaun Jun 28 '14 at 21:29

I was trying too hard.


When Goldblatt used the lattice equivalence $$x\sqsubseteq y \iff x=x\sqcap y$$ to write enter image description here as $$\{\langle x, y\rangle\vert x\smallfrown y=x\},$$ he did so for a reason: it's so that enter image description here is the set for which $\smallfrown$ and $pr_1$ agree, which is then, by $\S 3.10$, the equaliser of the two.


If you draw the commutative diagrams for the products $t_1=\langle\operatorname{true}_2=\operatorname{true}\circ\vert_2, 1_2\rangle$ and $t_2=\langle1_2,\operatorname{true}_2\rangle$ and write down all the equations (like $pr_2\circ t_1=1_2$), you'll see that $t_1$ and $t_2$ are nothing more than what Goldblatt said they were: $t_1$ sends $0$ to $\langle1, 0\rangle$ and $1$ to $\langle1, 1\rangle$; $t_2$, similarly; and we can identify them with $A$ and $B$, respectively, because $A$ and $B$ are two-element sets whose set-inclusions with respect to $2\times 2$ are $t_1$ and $t_2$, respectively.

The coproduct $2+2$ is just the disjoint union of $2$ with itself, so that we can write $$2+2=2^{(0)}\cup2^{(1)},$$ where $2^{(0)}=2\times\{0\}$ and $2^{(1)}=2\times\{1\}$, to see that $f$ is given by $$f(a, i)=\begin{cases}t_1(a) &: i=0 \\ t_{2}(a) &: i=1.\end{cases}$$

The rest is downhill from there (once you see that $\operatorname{im}f=\operatorname{im}t_1\cup\operatorname{im}t_2$).

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