equivalent characterisation of simply connect spaces

Question

I want to prove the following: Let $X$ be path connected space, $S^{1}$ the $1$-sphere and $D^{2}$ the unit circle. Following are equivalent: i)X is simply connected. ii) If $f:S^{1} \to X$ is continuous, then $f$ is homotopic to a constant function. iii) If $f:S^{1} \to X$ is continuous, then we can extend f to a continuous function from $D^{2}$ to $X$

My ideas for i) implies ii): Since X is simply connected, X has the trivial fundamental group $\pi_1(X)$. Thanks to $f:S^{1} \to X$, we have a group homomorphism $f^{*}:\pi_1(X) \to \pi_(S^{1}), [\phi] \mapsto [f\circ \phi]$. Since $\pi_1(X)$ is trivial, $f^{*}$ is the trivial group homomorphism. Can I now conclude that any loop in the image of f is homotopic to the constant loop? This would intuitivly imply that the image of f has no "holes". Probably helpful is also that, $\pi_(S^{1})\simeq \mathbb{Z}$.

The statement iii) seems to me quite far away from i) and ii).

Thanks for your answers.

Bests

bjn

Answer 1

The connections between i), ii) and iii) are that the domains of the various homotopies are closely related to each other via quotient maps.

Let me explain this by proving a couple of the directions.

Item ii) means that for any $f : S^1 \to X$ there exists a map $H : S^1 \times [0,1] \to X$ such that $h(x,0)=f(x)$ and $H | S^1 \times 1$ is constant.

To prove the equivalence of ii) and iii), one uses the fact that there is a quotient map $q : S^1 \times [0,1] \to D^2$ under which $S^1 \times 1$ is identified to the origin of $D^2$, and each segment $p \times [0,1] \subset S^1$ is mapped to the radial segment of $D^2$ going inward from $p$ to the origin. The formula for $q$ that accomplishes this is easily written down.

So if ii) holds then, since the homotopy $H : S^1 \times [0,1] \to X$ is constant on $S^1 \times 1$, one obtains an induced map on the quotient space of $S^1 \times [0,1]$ by collapsing $S^1 \times 1$ to a single point, and that quotient space is identified with $D^2$ via the map $q$. The map $H$ therefore induces the desired extension $F : D^2 \to X$.

And if iii) holds with extension $F : D^2 \to X$ then the composition $$H : S^1 \times [0,1] \xrightarrow{q} D^2 \xrightarrow{F} X $$ is the desired homotopy needed to prove ii).

Answer 2

Let $f_t$ be a homotopy with $f_0(\theta)=f(\theta)$, $f_1(\theta)=x\in X$. Let $G(r\cos\theta,r\sin\theta)=f_{1-r}(\theta)$. This is a continuous map from $D$ to $X$.