Surjectivity of Induced homomorphism on fundamental group

Question

Let X be a topological space, $x_0 \in X$ and $f: S^1 \rightarrow X$ be any continuous map. Show that the homomorphism $\pi_1 (X, x_0) \rightarrow \pi_1 (X \cup_f D^2, x_0)$ defined by the inclusion $\pi_1 (X, x_0) \hookrightarrow \pi_1 (X \cup_f D^2, x_0)$ is surjective.

Answer 1

Indeed, there is the said surjection:

a map from $\, S^1\, $ into $\,X\,$ with a disk attached can be homotopically moved in such a way that the image of $\,S^1\,$ will end up in $\,X\, $ (where $\, X\, $ is interpreted here as a subset of $\, X\,$ with a disk attached to it). And the points of the circle which were originally mapped into $\,X\, $ don't even have to move during the homotopy.

Here is a pretty detailed proof:

But first a remark,

Remark: We may assume that the base point $\,x_0\,$ belongs to $\,f(S^1).\,$ Then the case of arbitrary $\,x_0\,$ which belongs to $\,X\,$ follows (well, to the canonical image of $X$).   Furthermore, we may assume that the same $\,x_0\,$ is a base point of $\,S^1.$

Warning: There are points of $\,S^1\,$ just as (kind of abstract) $\,S^1.\, $ And there are points of $\,f(S^1).\,$ Keep them separated in your mind, there is nothing to worry.

Proof:

Let $\,\Bbb C\, $ be the field of the complex numbers, $\,D:=\{z\in\Bbb C: |z|\le 1\},\,$ $\,B:=\{z\in\Bbb C: |z|\le \frac12\},\,$ $\,S_B:=\{z\in\Bbb C: |z| = \frac12\},\,$ and $\,T:=f^{-1}(B).$

Let $\,g:S^1\to X\cup_f D\,$ be an arbitrary continuous loop for which $\,g(x_0)=x_0.$

There is continuous $\, h_0: g^{-1}(B)\to X\cup_f D\, $ such that $\,h_0(g^{-1}(B))\subseteq S_B\, $ and $\,h_0(x)=g(x)\,$ whenever $\,g(x)\in S_B\,$ -- this is so because $\,\dim(g^{-1}(B))=1.$

Also, $\,\, g|((X\cup_f D\setminus B)\cup S_B)\,:\, (X\cup_f D\setminus B)\cup S_B\, \to\, X\cup_f D\,\,$ is continuous.

Also, the join extension $\,\,h:S^1\to X\cup_f D\,\, $ of functions $\,\,g|((X\cup_f D\setminus B)\cup S_B)\,$ and $\,h_0\,\,$ is continuous. Obviously, $\,g\,$ and $\,h\,$ are homotopic under which $\,x_0\,$ is fixed all the time (from $0$ to $1$).

Finally, define $\,\ell:S^1\to X\cup_f D\,$,

$$ \forall_{x\in h^{-1}((X\cup_f D)\setminus(D\setminus S^1))} \quad \ell(x)=h(x) $$ and $$ \forall_{x\in D\setminus S^1}\quad \ell(x)\,=\, f\left(\frac {h(x)}{|h(x)|}\right) $$

(recall that above, $\,|h(x)|\ge\frac12.)$

Once again, $\,\ell\,$ and $\,h\,$ admit a homotopy under which $\,x_0\,$ is fixed. Thus the same is true for $\,g\,$ and $\,\ell.\,$ But

$$ \ell(S^1)\,\subseteq\,X $$

(assuming one of the equivalent constructions for which $\,X\subseteq X\cup_f D,\,\,$ i.e. $\, X\,=\,(X\cup_f D)\setminus(D\setminus S^1)\,\,) $

The theorem has been proven. Great!