Continued from last week...
Solution.
I’m going to start by taking a step back and looking at what I’m really trying to prove here. This exercise was worded using “if and only if”, so I need to prove both of the following statements:
- If an initial object is isomorphic to a terminal object, then the category has zero maps.
- If a category has zero maps, then an initial object is isomorphic to a terminal object.
In a category with initial and terminal objects, we’re assured to have exactly one map \(S \xrightarrow{m} T\text{.}\) This map is an isomorphism “if and only if” there’s a respective inverse map \(T \xrightarrow{m^{-1}} S\) such that \(m^{-1} \circ m = 1_S\) and \(m \circ m^{-1} = 1_T\text{.}\)
We’ve defined “zero maps” by the property that any map composed with a zero map produces another zero map. In symbolic form:
\begin{equation*}
\forall A \xrightarrow{f} B, B \xrightarrow{g} C:
\text{zero_map}(f) \lor \text{zero_map}(g) \implies
\text{zero_map}(g \circ f)
\end{equation*}
Okay, lets focus on the first statement. We assume \(m^{-1}\) exists then need to show that it gives rise to some “zero map”. Specifically, I think we can define \(X \xrightarrow{0_{XY}} Y\) relative to the uniquely defined maps \(X \xrightarrow{t_x} T\) and \(S \xrightarrow{s_y} Y\text{.}\) Simply define \(0_{XY}\) as the composition \(X \xrightarrow{t_x} T \xrightarrow{m^{-1}}
S \xrightarrow{s_y} Y
\text{.}\)
So how can we demonstrate that \(0_{XY}\) is a “zero map”? We’d need to look at every possible composition \(A \xrightarrow{f} X
\xrightarrow{0_{XY}} Y\) and \(X \xrightarrow{0_{XY}} Y \xrightarrow{g} B\) to ensure we get another zero map.
We can define \(A \xrightarrow{0_{AY}} Y\) relative to the unique maps \(A \xrightarrow{t_a} T\) and \(S \xrightarrow{s_y} Y\) by \(A \xrightarrow{t_a} T \xrightarrow{m^{-1}}
S \xrightarrow{s_y} Y
\text{.}\) If \(0_{XY} \circ f \neq 0_{AY}\text{,}\) there would need to be some point \(T \xrightarrow{a} A\) such that \(0_{XY} \circ f \circ a \neq 0_{AY} \circ a\text{.}\) Expanding with our definitions of the zero maps:
\begin{equation*}
s_y \circ m^{-1} \circ t_x \circ f \circ a
\neq
s_y \circ m^{-1} \circ t_a \circ a
\end{equation*}
Which can only happen if
\begin{equation*}
t_x \circ f \neq t_a
\end{equation*}
But \(A \xrightarrow{t_a} T\) is a uniquely defined map with that codomain and domain! It follows from this contradiction that our assumption \(0_{XY} \circ f \neq 0_{AY}\) is false, and consequently that \(0_{XY} \circ f = 0_{AY}\) is a zero map.
Likewise, we can define \(X \xrightarrow{0_{XB}} B\) relative to the unique maps \(X \xrightarrow{t_x} T\) and \(S \xrightarrow{s_b} B\) by \(X \xrightarrow{t_a} T \xrightarrow{m^{-1}}
S \xrightarrow{s_b} B
\text{.}\) If we assume \(g \circ 0_{XY} \neq 0_{XB}\text{,}\) then there must exist some \(T \xrightarrow{x} X\) satisfying \(g \circ 0_{XY} \circ x \neq 0_{XB} x\text{.}\) Expanding this gives us the following:
\begin{equation*}
g \circ s_y \circ m^{-1} \circ t_x \circ x
\neq s_b \circ m^{-1} \circ t_x \circ x
\end{equation*}
Which can only happen when:
\begin{equation*}
g \circ s_y \neq s_b
\end{equation*}
Since \(s_b\) is the unique map with that domain and codomain, we get a contradiction. Our assumption that \(g \circ 0_{XY} \neq 0_{XB}\) must therefore be false.
Essentially, by having a map \(T \xrightarrow{m^{-1}} S\) we can define a zero map \(0_{AB}\) between any objects \(A,B\) in \(\mathcal{C}\) through the unique composition \(A \rightarrow T \xrightarrow{m^{-1}} S \rightarrow B\text{.}\)
Having demonstrated that we can construct our zero maps from \(m^{-1}\text{,}\) let’s attempt to do the reverse.
Suppose we have a zero maps \(0_{XY},0_{AY},0_{XB}\) with the property that for any maps \(A \xrightarrow{f} X\) and \(Y \xrightarrow{g} B\text{,}\) the composition \(0_{XY} \circ f = 0_{AY}\) and \(g \circ 0_{XY} = 0_{XB}\text{.}\) The associative property of the category says that \(g \circ (0_{XY} \circ f) = (g \circ 0_{XY}) \circ f\text{,}\) so we must have the case that \(g \circ 0_{AY} = 0_{XB} \circ f = 0_{AB}\text{.}\)
If this is true for any maps \(f,g\text{,}\) let’s choose the uniquely defined maps \(f = S \xrightarrow{j_1} (S+T)\) and \(g = (S \times T) \xrightarrow{p_2} T\text{.}\) It follows from the above that \(g \circ 0_{S(S+T)} = 0_{X(S \times T)} \circ f = 0_{ST} \text{.}\) Let’s see we can assemble this into the commutative diagram from last week:
In this diagram, we get a zero map from \(T \rightarrow S\) defined through this unique composition \(p_1 \circ p \circ j \circ j_2\text{.}\) This map has a inverse given by \(p_2 \circ p \circ j \circ j_1\text{.}\) Knowing that \(j \circ j_2 = 1_T\) and \(p_2 \circ p = 1_T\text{,}\) we could simplify these maps down to \(p_1 \circ p = 0_{ST}\) and \(j \circ j_1 = 0_{TS}\text{.}\) These maps satisfy the property of the \(m^{-1}\) we were looking for, namely that \(0_{ST} \circ 0_{TS} = 1_T\) and \(0_{TS} \circ 0_{ST} = 1_S\text{.}\)
With that, I believe this exercise is complete.
