Fundamental groups and some properties

Question

I have some basic questions about fundamental groups that came up when I tried to prove a few things: I am sorry that they are kind of informal questions, but I could not find any answers to them in my texbook and I think it may be good to know these kind of things.

If I have a set $A$ and a set $B$, such that $A \subset B$, does the canonical embedding always mean that there is a homomorphism $\pi_1(A) \rightarrow \pi_1(B)$ between the fundamental groups?

In principle fundamental groups for a set are defined with respect to a point. What is the exact condition under which you can talk about the fundamental group of a set rather than of a particular point? I guess it is sufficient that the set is connected, but maybe path-connected is necessary?

If I have a disc $D^2 \subset \mathbb{R}^2$ and want to find a surjective continuous map $f$ into the sphere $S^1$. Can such a map exist? The thing is that I was wondering whether this may result in problems with the fundamental groups: $\pi_1(D^2)=0$ and $\pi_1(S^1)= \mathbb{Z}$. So, if $f$ is a continuous map with a special property like being (surjective, injective, bijective, etc., a homeomorphism) do any of these kind of property result in nice properties of the homomorphism?

Answer 1

Any function $f : X \to Y$ induces a homomorphism on fundamental groups $f_* : \pi_1(X, x_0) \to \pi_1(Y, f(x_0))$. If a space is path-connected, the fundamental group is unaffected by the choice of the base point. This is why the base point is sometimes omitted.

In general, injectivity and surjectivity are not preserved when passing to homomorphisms.

Consider the inclusion $S^1 \hookrightarrow \Bbb R^2$. This map is injective, but the induced homomorphism sends $\pi_1(S^1) \cong \Bbb Z$ to the trivial group; hence it cannot be injective.

Consider the map $S^1 \to S^1$, $z \mapsto z^2$. This map is surjective but the induced homomorphism is multiplication by $2$. Hence it's not surjective.

However, passing to homomorphisms respects function composition. In other words, if $h = g \circ f$, then $h_* = g_* \circ f_*$. This is used to show that no retraction exists from $D^2$ to $S^1$ as you hint in your question. Let $i : S^1 \to D^2$ be the inclusion, and $r : D^2 \to S^1$ be a retraction. Since $\Bbb 1_{S^1} = r \circ i$, the same is true for induced homomorphisms: $\Bbb 1_{S^1*} = r_* \circ i_*$. This means that $i_*$ is injective and $r_*$ is surjective. At the same time, $r_* : \pi_1(D^2) \to \pi_1(S^1)$ is trivial since $\pi_1(D^2)$ is trivial. This results in a contradiction.

Answer 2

The fundamental group $\pi_1$ is a functor $\mathsf{Top}_*\to\mathsf{Grp}$. Here $\mathsf{Top}_*$ is the category of based topological spaces whose objects are pairs $(X,x)$ where $X$ is a topological space and $x$ is a point of $X$. The morphisms $(X,x)\to(Y,y)$ in $\mathsf{Top}_*$ are the continuous maps $f:X\to Y$ such that $f(x)=y$. The category $\mathsf{Grp}$ is the category of groups and group homomorphisms.

Now, the fact that $\pi_1$ is a functor $\mathsf{Top}_*\to\mathsf{Grp}$ means it assigns a group $\pi_1(X,x)$ to every based topological space $(X,x)$ and it assigns a group homomorphism $\pi_1(f):\pi_1(X,x)\to\pi_1(Y,x)$ to every morphism $f:(X,x)\to(Y,y)$. In particular, an inclusion $i=A\hookrightarrow X$ of a subspace may be viewed as a morphism $i:(A,x)\to(X,x)$ where $x$ is any point of $A$. The discussion above tells us that $i$ induces a group homomorphism $\pi_1(i):\pi_1(A,x)\to\pi_1(X,x)$.

Another nice property of functors is that they preserve identities and composition. That is $\pi_1(\DeclareMathOperator{id}{id}\id_{(X,x)})=\id_{\pi_1(X,x)}$ and $\pi_1(f\circ g)=\pi_1(f)\circ\pi_1(g)$. What this really means is that $\pi_1$ maps commutative diagrams to commutative diagrams.

Thus if $f:D^2\to S^1$ has a continuous right inverse $r:S^1\to D^2$, then we have a commutative diagram $$ \begin{array}{ccc} S^1 & \xrightarrow{r} & D^2 \\ _{\id_{S^1}}\downarrow & & \downarrow_{f} \\ S^1 & \xrightarrow{\id_{S^1}} & S^1 \end{array} $$ where the base-point notation is suppressed. Applying $\pi_1$ to the diagram gives a commutative diagram of groups and group homomorphisms $$ \begin{array}{ccc} \pi_1(S^1) & \xrightarrow{\pi_1(r)} & \pi_1(D^2) \\ _{\pi_1(\id_{S^1})}\downarrow & & \downarrow_{\pi_1(f)} \\ \pi_1(S^1) & \xrightarrow{\pi_1(\id_{S^1})} & \pi_1(S^1) \end{array} $$ Using that $\pi_1(S^1)\simeq\Bbb Z$, that $\pi_1(D^2)\simeq 0$, and that $\pi_1$ preserves identities allows us to rewrite the diagram as $$ \begin{array}{ccc} \Bbb Z & \xrightarrow{\pi_1(r)} & 0 \\ _{\id_{\Bbb Z}}\downarrow & & \downarrow_{\pi_1(f)} \\ \Bbb Z & \xrightarrow{id_{\Bbb Z}} & \Bbb Z \end{array} $$ But this gives $\pi_1(f)\circ\pi_1(r)=0$ and $\pi_1(f)\circ\pi_1(r)=\id_{\Bbb Z}$ which is impossible. Hence no right inverse of $f:D^2\to S^1$ exists.