Results on constant morphisms

Lemma 1.

A constant morphism in $\Set$ is the same as a constant map in the usual sense.

Proof. Let $X,Y$ be two sets and let $f : X \to Y$ be a map. If $f$ is a constant morphism, then, in particular, for all morphisms $x_1,x_2 : 1 \to X$ , we have $f \circ x_1 = f \circ x_2$ . Therefore, $f(x_1) = f(x_2)$ for all $x_1,x_2 \in X$ , so $f$ is a constant map. Conversely, if $f$ is a constant map and $x_1,x_2 : T \rightrightarrows X$ are two maps, then $f \circ x_1 = f \circ x_2$ , since for all $t \in T$ we have $f(x_1(t)) = f(x_2(t))$ . $\square$

Lemma 2.

Let $\C$ be a locally small category. Any representable functor $U : \C \to \Set$ maps constant morphisms to constant maps.

Proof. We may assume that $U = \Hom(G,-)$ for some $G \in \C$ . If $f : X \to Y$ is a constant morphism in $\C$ , then the induced map $f_* : \Hom(G,X) \to \Hom(G,Y)$ is constant by the definition of a constant morphism. $\square$

Lemma 3.

Any right adjoint functor preserves constant morphisms.

Proof. Let $G : \C \to \D$ be a functor that is right adjoint to $F : \D \to \C$ . Let $f : X \to Y$ be a constant morphism in $\C$ . To show that $G(f) : G(X) \to G(Y)$ is constant, let $y_1,y_2 : T \rightrightarrows G(X)$ be two morphisms. Under the adjunction, these correspond to morphisms $x_1,x_2 : F(T) \rightrightarrows X$ . Since $f$ is constant, we have $f \circ x_1 = f \circ x_2$ as morphisms $F(T) \rightrightarrows Y$ . Hence, $G(f) \circ y_1 = G(f) \circ y_2$ as morphisms $T \rightrightarrows G(Y)$ . $\square$

Lemma 4.

If $X$ is a subterminal object, then any morphism $X \to Y$ is constant. If $Y$ is a terminal object, then any morphism $X \to Y$ is constant.

Proof. This is immediate from the definitions. $\square$

Lemma 5.

If $f : X \to Y$ is a monomorphism that is constant, then $X$ is subterminal.

Proof. If $x_1,x_2 : T \rightrightarrows X$ are morphisms, then $f \circ x_1 = f \circ x_2$ since $f$ is constant. Since $f$ is also a monomorphism, we infer that $x_1 = x_2$ . $\square$

Of course, all results on constant morphisms dualize to results on coconstant morphisms (except for Lemma 2). For example, Lemma 5 implies that if an epimorphism $f : X \to Y$ is coconstant, then $Y$ is "co-subterminal", i.e. every two morphisms $Y \rightrightarrows T$ are equal.

Context

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