# Show that Hom(X, -) is always a functor

I'm totally new to the concept of functors. All I know about them is the fact that when they get an object as argument, it acts like an object, and when it takes a morohism, it acts as a morphism.

I'm quite new to the Hom function too, I know that Hom(A, B) is the set of all morphisms which domain is A and codomain is B.

I've been asked to prove that Hom(X,-) is always a functor, but I don't know how to start, since I don't have any clear properties in my hands.

• In which category? Sep 18, 2016 at 20:37
• Pos (posets) category Sep 18, 2016 at 20:51

For a functor we have to give a definition of both objects and morphisms, so define for any category $\mathcal C$ : $Hom_{\mathcal C}(X,\_): \mathcal C \to (Set)$ $Hom_{\mathcal C}(X,\_)(Y) := Hom_{\mathcal C}(X,Y)$
And for a morphism $f:Y\to Z$ $Hom_{\mathcal C}(X,\_)(f):= ( h \mapsto (f \circ h))$
Lastly we have to check, that for maps $f:W\to Y$ and $g:Y\to Z$, $Hom_{\mathcal C}(X,\_)(f) \circ Hom_{\mathcal C}(X,\_)(g) = Hom_{\mathcal C}(X,\_)(f\circ g)$, but this is clear as $(f\circ g)\circ h = f \circ (g \circ h )$ by the definition of a category.