![homfunctor_opposite_category](http://aether.co.kr/images/homfunctor_op.svg)

So I think I got why we need the opposite category to send pairs of objects to homsets. As seen in the picture above in order to get from \\(\mathcal{C}(a,b)\\) to \\(\mathcal{C}(a',b)\\), we need to precompose f so that the arrows compose.

Now I have a question about how the homfunctor preserves composition. I will use the same example with two objects and one non-trivial morphism above to pose the question.

If we have two composable morphisms \\( \mathcal{C}(f,1_{b'}) \circ \mathcal{C}(1_a,g) = \mathcal{C}(f,g)\\) , the functor needs to preserve this via \\( \mathcal{C}(f \circ 1_a,1_{b'} \circ g)\\). But the order doesn't seem right in that \\(f \circ 1_a\\) and \\(1_{b'} \circ g\\) are both not composable. What am I doing wrong here?

Edit: Oops \\(1_{b'} \circ g\\) does compose.

So I think I got why we need the opposite category to send pairs of objects to homsets. As seen in the picture above in order to get from \\(\mathcal{C}(a,b)\\) to \\(\mathcal{C}(a',b)\\), we need to precompose f so that the arrows compose.

Now I have a question about how the homfunctor preserves composition. I will use the same example with two objects and one non-trivial morphism above to pose the question.

If we have two composable morphisms \\( \mathcal{C}(f,1_{b'}) \circ \mathcal{C}(1_a,g) = \mathcal{C}(f,g)\\) , the functor needs to preserve this via \\( \mathcal{C}(f \circ 1_a,1_{b'} \circ g)\\). But the order doesn't seem right in that \\(f \circ 1_a\\) and \\(1_{b'} \circ g\\) are both not composable. What am I doing wrong here?

Edit: Oops \\(1_{b'} \circ g\\) does compose.