Question about simply connectedness

Question

Let $X$ be a path connected space. I have to prove the following:

If each $f \in \mathcal{C}(\mathbb{S}^1;X)$ is homotopic to a constant map, then $\pi_1(X)$ is trivial ($X$ is simply connected) .

I am trying to prove by using this result:

Let $(Y,\mathcal{T}_Y)$ and $(Z,\mathcal{T}_Z)$ topological spaces and $\sim$ a equivalence relation on $Y$. If $f\in \mathcal{C}(Y;Z)$ then the map $\tilde{f}:Y/_\sim \hookrightarrow Z, \tilde{f}[y]=f(y)$ is continous and $f=\tilde{f} \circ p$ where $p$ is the canonical quotient projection.

So, in order to prove it denote $I:=[0,1]$ and let $x_0 \in X$ and $\alpha \in \mathcal{C}(I;X):\alpha(0)=\alpha(1)=x_0$ . Consider $\sim$ the equivalence relation such that $$x \sim y \iff x=y \lor x,y \in \{0,1\} $$

It is well known that with this relation, we can identify $I/_\sim$ with $\Bbb{S}^1$, so we can see the corresponding $\tilde{\alpha}$ as a map from $\Bbb{S}^1$ to $X$. Then, by the hypothesis, we have that $\tilde{\alpha}$ is homotopic to a constant map. In other words: $$\exists \tilde{H} \in \mathcal{C}(\Bbb{S}^1 \times I ; X), \forall x \in \Bbb{S}^1 : \tilde{H}(x,0) = \tilde{\alpha}(x) \land \tilde{H}(x,1) = c \in X$$

Since every couple of constant maps are homotopic, we can take $x_0$ as $c$ .

Consider $H : I \times I \hookrightarrow X, (t,s) \mapsto \tilde{H}(p(t),s)$ which is clearly contiunous and $$\forall t \in I : H(t,0) = \tilde{H}(p(t),0) = \tilde{\alpha}(p(t)) = \alpha(t) \land H(t,1) = x_0$$

The problem is that I do not know how I can claim that for all $s \in I: H(0,s)=x_0=H(1,s)$ in order to say that $H$ is a path homotopy, so any possible help would be appreciated.

EDIT

I think that it can be solved if we used this result:

Given $X,Y$ topological spaces and $f,g \in \mathcal{C}(X;Y)$ with $\varphi$ an homotopy between them, the induced homomorphisms $f_{\ast}, g_{\ast}$ satisfies $f_{\ast} = \beta_h \circ g_{\ast}$ where $h = \varphi (x_0, -)$ for some fixed $x_0 \in X$ and $\beta_h = [h] \ast - \ast [h']$ where $h'$ denotes the inverse path of $h$ .

I proved that $\tilde{\alpha}$ is homotopic to $k_{x_0}$ ( the map which assings $x_0$ to each $p\in \Bbb{S}^1$ ). Then, by the result :

$$\tilde{\alpha}_{\ast} = \beta_h \circ k_{x_0 \ast} \Rightarrow \forall \gamma \in \pi_1(\Bbb{S}^1) : \tilde{\alpha}_{\ast} ([\gamma]) = \beta_h \circ k_{x_0 \ast} ([\gamma]) = \beta_h ([c_{x_0}]) = [h \ast c_{x_0} \ast h' ] = [c_{x_0}]$$

Then, $\tilde{\alpha}_{\ast}$ is the trivial homomorphism. Now, note that $p : I \hookrightarrow I/_\sim$ is in fact a path on $\Bbb{S}^1$ via the identification $\Bbb{S}^1 = I/_\sim$. Therefore, we have:

$$[\alpha] = [\tilde{\alpha} \circ p] = \tilde{\alpha}_{\ast} ([p]) = [c_{x_0}]$$

and this conclude the proof. Is this fine?

Answer 1

Yes, this is true. The linked post (or rather, the author of the textbook) made the critical error of stating the result with the unit interval. Trivially, any map $I\to X$ is homotopic to any other so the result erroneously claims every loop is homotopic to every other...

But if you are more careful and work with $S^1$, which I expect the author had intended, then it is true that "all maps $S^1\to X$ are homotopic iff. $X$ is simply connected". Warning: this is a nice and a surprising thing; it is not true in general that "all maps $Y\to X$ are nullhomotopic" implies "all maps $Y\to X$ are nullhomotopic rel. $\{y\}$".

This would follow from the more important result:

If $f:S^n\to X$ is nullhomotopic, then it is nullhomotopic rel. $\{s\}$ for any fixed $s$.

The proof - well, I'm sure this features elsewhere on the site, I think I've even written it myself before - goes as follows:

Note $D^{n+1}$ "is" the quotient of $S^n\times I/(S^n\times\{0\})$ (polar coordinates). Thus a nullhomotopy of $f$ (not necessarily relative to anything) is "just" a map $G:D^{n+1}\to X$ such that $G(x)=f(x/\|x\|)$ for $x$ on the boundary sphere.

So, we are given such a $G$. Then we use a clever squashing map, that is, a quotient $\Gamma:D^{n+1}\twoheadrightarrow D^{n+1}$ which exhibits $D^{n+1}$ as the reduced cone of $S^n$. Specifically $\Gamma(r,\theta)=r\cdot\theta+(1-r)\cdot s$ in polar coordinates (this is the convex combination connecting a point $(r,\theta)$ to the point $s$ on the boundary sphere). $\Gamma$ fixes the boundary and computes the quotient $D^{n+1}/[0,s]\cong D^{n+1}$.

The composite $G\circ\Gamma:D^{n+1}\to X$ then induces a nullhomotopy of $f$ with now the special property that it is constant on $\{s\}\times I$. That is, $f$ is nullhomotopic rel $\{s\}$ which is exactly what we want.

Since $\pi_1(X,x_0)$ "is" the homotopy classes of maps $(S^1,s)\to(X,x_0)$ rel. $s$ (for any $s$, it doesn't matter - up to isomorphism) we get the original claim.