Confusion about the underlying function of a Functor.

Question

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.

Answer 1

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.

This is a feature, not a bug - we definitely want to view putting the circle inside itself as distinct from putting the circle into the plane, especially from the homotopic point of view.

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.