Our professor has used the fact that the image of a nullhomotopic $\varphi \simeq 0$ is contractible. While this seems very intuitive, I decided I want to proove this and it turned to be harder than I thought. I tried this:
So let $\varphi \simeq 0$, so we have a homotopy $H: X \times I \rightarrow X$ such that $H_0 = \varphi$ and $H_1 = 0$ where $0$ is a constant map. We need to show that $im(\varphi) \simeq \{ * \} $, so we need maps $f: im(\varphi) \rightarrow \{ * \} $ and $g: \{ * \} \rightarrow im(\varphi)$ such that $f \circ g \simeq id_{\{ * \}} $ and $g \circ f \simeq id_{im(\varphi)}$. First we note that $g\circ f = id$, so we only need to make sure $f \circ g \simeq id_{\{ * \}}$. Set $z = g(*) \in im(\varphi)$, so $z=f(y)$ for some $y \in X$. So we have $(f \circ g)(x)=z$. We need to find a homotopy $h$ such that $h_0=f \circ g$ and $h_1=id$. We find that by setting $h(x,t)=H(y,t)$ because then $h(x,0)=H(y,0)=\varphi(y)=z=(f \circ g)(x)$ and $h(x,1)=H(x,1)=0$.
But I get the constant map, not the identity. I'm stuck on this for hours. How to proove this?
Best Answer
The statement is false. Take the image of the inclusion of $S^1$ inside $D^2$, which is a nullhomotopic map. The image is homeomorphic to $S^1$ which is not contractible.
Even if you suppose that both the source and the target are contractible, this does not hold : take the image of the exponential map $t \mapsto e^{it}$ from $\mathbb R$ to $\mathbb R^2$.