# Confusion about the underlying function of a Functor.

I was reading about Brower's fixed point theorem and a doubt came to mind about the underlying function of a functor on morphisms. We can think of $$\mathbb{S}^1$$ as a subset of $$\mathbb{R}^2$$, so we get the inclusion map $$i : \mathbb{S}^1 \hookrightarrow \mathbb{R}^2$$, of course we have the identity map $$\mathrm{id}: \mathbb{S}^1 \rightarrow \mathbb{S}^1$$. As we know, the first fundamental group is a functor so $$\pi^1(i : \mathbb{S}^1 \hookrightarrow \mathbb{R}^2)=0$$ since $$\mathbb{R}^2$$ is contractible, also $$\pi^1(\mathrm{id}: \mathbb{S}^1 \rightarrow \mathbb{S}^1)= \mathrm{id}_\mathbb{Z}: \mathbb{Z}\rightarrow \mathbb{Z}$$. So far so good, but as functions, i.e. some subset of a cartesian product with some property, $$i= \mathrm{id}$$, and thus the underlying function of the functor is ill defined. So, what I'm asking is, what am I missing? Are these functions different since we are looking them as morphisms? From what definition does this follow? I'm pretty sure that when we say that a morphism has a domain and a codomain we actually are saying the morphisms are triples, but I'm feeling little insecure. Any insight is appreciated.

A continuous map $$f:(X,x_0)\to (Y,y_0)$$ induces a homomorphism $$f_*:\pi_1(X,x_0)\to\pi_1(Y,y_0)$$ between the fundamental groups, two examples being the identity and inclusion maps $$f:S^1\to S^1, \ g:S^1\to \mathbb{R}^2$$ which give rise to the identity and trivial homomorphisms $$f_*:\mathbb{Z}\to\mathbb{Z}, \ g_*:\mathbb{Z}\to1.$$ The two "morphisms" $$f$$ and $$g$$ are not the same, since they have different codomains.
In general, a (covariant) functor $$\mathcal{F}$$ takes objects $$A$$ and $$B$$ to objects $$\mathcal{F}(A)$$ and $$\mathcal{F}(B)$$ and a morphism $$f:A\to B$$ to the morphism $$\mathcal{F}(f):\mathcal{F}(A)\to\mathcal{F}(B)$$ (respecting composition, etc.). In other words, the codomain is part of the information of a morphism as you say.
For the fixed-point theorem, the assumption of a retract $$r:D\to S^1$$ from the disc to its boundary circle gives a contradiction after applying the $$\pi_1$$ functor. The composition $$S^1\xrightarrow{i}D\xrightarrow{r}S^1$$ becomes $$\mathbb{Z}\xrightarrow{i_*}1\xrightarrow{r_*}\mathbb{Z}.$$ On the one hand $$\pi_1(r\circ i)$$ should be the identity homomorphism since $$r\circ i$$ is the identity map. However, this homomorphism factors through the trivial group and must therefore be trivial. Hence the retract $$r$$ cannot exist.