L&S Cat Theory: Attempted solutions to "Conceptual Mathematics" by Lawvere and Schanuel

Given \(C\) such that for every \(X\) there is at most one map \(X \longrightarrow C\text{,}\) prove is a product.

Given \(C\) such that for every \(X\) there is at most one map \(C \longrightarrow X\text{,}\) prove is a sum.

For \(C, C \xrightarrow{p_1 = 1_C} C, C \xrightarrow{p_2 = 1_C} C\) to be a product, we need to prove that for any object \(X\) and pair of maps \(X \xrightarrow{f_1} C\) and \(X \xrightarrow{f_2} C\text{,}\) we have exactly one map \(X \xrightarrow{f} C\) for which \(f_1 = p_1 f\) and \(f_2 = p_2 f\text{.}\)

For \(C, C \xrightarrow{j_1 = 1_C} C, C \xrightarrow{j_2 = 1_C} C\) to be a sum, we need to prove that for any object \(X\) and pair of maps \(C \xrightarrow{g_1} X\) and \(C \xrightarrow{g_2} X\text{,}\) we have exactly one map \(C \xrightarrow{g} X\) for which \(g_1 = g j_1\) and \(f_2 = g j_1\text{.}\)

Let \(X \xrightarrow{f_1} C\) and \(X \xrightarrow{f_2} C\) be an arbitrary pair of maps map in the category.

Let \(C \xrightarrow{g_1} X\) and \(C \xrightarrow{g_2} X\) be an arbitrary pair of maps map in the category.

We’ve defined \(C\) such there is “at most” one map \(X \xrightarrow{f} C\text{.}\) Given that \(f_1,f_2\) exist, they must both be equal to this unique map \(f\text{.}\)

We’ve defined \(C\) such there is “at most” one map \(C \xrightarrow{g} X\text{.}\) Given that \(g_1,g_2\) exist, they must both be equal to this unique map \(g\text{.}\)

Since \(f = f_1 = f_2\) and \(p_1 = p_2 = 1_C\text{,}\) \(f_1 = p_1 f\) and \(f_2 = p_2 f\) both become the composition \(f = 1_C f\text{.}\) This follows directly from the definition of the identity map, so we’ve effectively demonstrated that \(C,1_C,1_C\) is a product.

Since \(g = g_1 = g_2\) and \(j_1 = j_2 = 1_C\text{,}\) \(g_1 = g j_1\) and \(g_2 = g j_2\) both become the composition \(g = g 1_C\text{.}\) This follows directly from the definition of the identity map, so we’ve effectively demonstrated that \(C,1_C,1_C\) is a sum.

Show that [the above property] is equivalent to the statement that the unique map \(C \longrightarrow \mathbf{1}\) is a monomorphism.

Show that [the above property] is equivalent to the statement that the unique map \(\mathbf{0} \longrightarrow C\) is a epimorphism.

A monomorphism \(C \xrightarrow{r} \mathbf{1}\) is defined as an “injective map” such that for every pair of maps \(\mathbf{1} \xrightarrow{c_1} C\) and \(\mathbf{1} \xrightarrow{c_2} C\text{,}\) \(r \circ c_1 = r \circ c_2\) implies \(c_1 = c_2\text{.}\)

A epimorphism \(\mathbf{0} \xrightarrow{s} C\) is defined as an “surjective map” such that for every pair of maps \(\mathbf{1} \xrightarrow{c_1} C\) and \(\mathbf{1} \xrightarrow{c_2} C\text{,}\) \(c_1 \circ s = c_2 \circ s\) implies \(c_1 = c_2\text{.}\)

Consider the composition defined by \(r \circ f: X \rightarrow \mathbf{1}\text{.}\) By the definition of terminal object, there is exactly one map \(X \longrightarrow \mathbf{1} \text{,}\) so \(r \circ f\) must be this unique map. Thus, having \(f\) with the “at most one map” property implies \(r\) is a monomorphism.

Consider the composition defined by \(g \circ s: \mathbf{0} \rightarrow X\text{.}\) By the definition of initial object, there is exactly one map \(\mathbf{0} \longrightarrow X\text{,}\) so \(g \circ s\) must be this unique map. Thus, having \(g\) with the “at most one map” property implies \(s\) is a epimorphism.

Next, we must prove the converse. We start by supposing \(r\) is a “monomorphism”.

Next, we must prove the converse. We start by supposing \(s\) is a “epimorphism”.

The statement \(r \circ c_1 = r \circ c_2 \implies c_1 = c_2\) is logically equivalent to the contrapositive \(c_1 \neq c_2 \implies r \circ c_1 \neq r \circ c_2\text{.}\) However, \(r \circ c_1\) and \(r \circ c_2\) are both defined with domain and codomains of \(\mathbf{1} \longrightarrow \mathbf{1}\) and the definition of terminal objects says this map is unique.

The statement \(c_1 \circ s = c_2 \circ s \implies c_1 = c_2\) is logically equivalent to the contrapositive \(c_1 \neq c_2 \implies c_1 \circ s \neq c_2 \circ s\text{.}\) However, \(c_1 \circ s\) and \(c_2 \circ s\) are both defined with domain and codomains of \(\mathbf{0} \longrightarrow \mathbf{1}\) and the definition of initial objects says this map is unique.

By assuming there are two maps \(X \xrightarrow{f_1} C\) and \(X \xrightarrow{f_2} C\) with \(f_1 \neq f_2\text{,}\) there would need to exist some a map \(\mathbf{1} \xrightarrow{x} X\) such that \(f_1 x \neq f_2 x\text{.}\) However, this means the compositions \(X \xrightarrow{f_1} C \xrightarrow{r} \mathbf{1}\) and \(X \xrightarrow{f_2} C \xrightarrow{r} \mathbf{1}\) are both maps \(X \rightarrow \mathbf{1}\text{.}\) By the definition of terminal object, there is only one such map \(X \rightarrow \mathbf{1}\) so we’re forced to conclude that \(r f_1 = r f_2\) are both equivalent to this unique map.

By assuming there are two maps \(C \xrightarrow{g_1} X\) and \(C \xrightarrow{g_2} X\) with \(g_1 \neq g_2\text{,}\) there would need to exist some a map \(\mathbf{1} \xrightarrow{c} C\) such that \(g_1 c \neq g_2 c\text{.}\) However, this means the compositions \(\mathbf{0} \xrightarrow{s} C \xrightarrow{g_1} X\) and \(\mathbf{0} \xrightarrow{s} C \xrightarrow{g_2} X\) are both maps \(\mathbf{0} \rightarrow X\text{.}\) By the definition of initial object, there is only one such map \(\mathbf{0} \rightarrow X\) so we’re forced to conclude that \(g_1 s = g_2 s\) are both equivalent to this unique map.

Effectively, if we have a monomorphism \(r: C \rightarrow \mathbf{1}\) we can construct the unique map \(X \xrightarrow{f} C\) relative to \(r^{-1}: \mathbf{1} \rightarrow C\) by way of the composition \(X \rightarrow \mathbf{1} \xrightarrow{r^{-1}} C\text{.}\) If such an inverse exists it must necessarily be unique, so the “at most one map” property follows from this uniqueness.

Effectively, if we have a epimorphism \(s: \mathbf{0} \rightarrow C\) we can construct the unique map \(C \xrightarrow{g} X\) relative to \(s^{-1}: C \rightarrow \mathbf{0}\) by way of the composition \(C \xrightarrow{s^{-1}} \mathbf{0} \rightarrow X\text{.}\) If such an inverse exists it must necessarily be unique, so the “at most one map” property follows from this uniqueness.