I am taking an course in Algebraic topology, and unfortunately, I'm unable to understand some proofs.
So, I am asking it here.
Prove that following conditions on a space $X$ are equivalent:
(i) The identity map of $X$ is null homotopic.
(ii) For every $Y$, every map $h: Y \to X$ is null homotopic.
So, for (i) $\Rightarrow$ (ii), let $i: X\rightarrow X$ is identity map and is given homotopic to $c$, which implies there exists an homotopy $\{h_t\}$ such that $h_0 = i$ and $h_1= c$. But how to show that for every $Y$, every map $h:Y \rightarrow X$ is null homotopic.
Converse , I proved taking $Y=X$ in (ii).
So, can you please help me with (i) $\Rightarrow$ (ii)?
Best Answer
Here is a hint spelling out the desired nullhomotopy in detail.
Let $F:X \times I \to X$ be a homotopy with $F \mid_{X \times \{0\}}=Id_X$ and $F \mid_{X \times\{1\}}=c_x$, where $c_x:X \to X$ is a constant map $c_x(z)=x$ for all $z \in X$.
Let $g:Y \to X$ be an arbitrary continuous map. Then define $G:Y \times I \to X$ by $G(y,t)=F(g(y),t)$.
This is a nullhomotopy from $g$ to a constant map.