# Hatcher 0.10

Hatcher, Algebraic Topology, Chapter 0

10. Show that a space $X$ is contractible if and only if every map $f:X\to Y$, for arbitrary $Y$, is null-homotopic. Similarly, show $X$ is contractible if and only if every map $f:Y\to X$ is null-homotopic.

Proof. First, recall what the terms in this problem mean:

• A map $f:X\to Y$ is null-homotopic if it is homotopic to a constant map.
• A space $X$ is contractible if it is homotopy equivalent to a point $\{x_0\}$, a useful equivalent of which is to say that $X$ is contractible whenever the identity map $id_X:X\to X$ is homotopic to a constant map.

Throughout, let $g:X\to x_0$ and $h:x_0\to X$ be such that $gh\simeq 1_{x_0}$ and $hg\simeq 1_X$. For convenience, consider the first claim to be part (a) and the second part (b).

(a) $\left(\Longrightarrow\right)$ Suppose that $X$ is contractible and let $f:X\to Y$ be a map for $Y$ arbitrary. Intuitively, $X$ homotopy equivalent to $x_0$ implies that $X$ can be “continuously deformed” to $x_0$, whereby it seems logical that the map $f:X\to Y$ can be likewise “continuously deformed” into a map $\widehat{f}:x_0\to Y$, that is, a constant map. To be precise, define $\widehat{f}=f\circ h:x_0\to Y$ where $h$ is the half of the homotopy equivalence mentioned above. Clearly, then, $\widehat{f}$ is constant, and because $\widehat{f}\circ g = (f\circ h)\circ g\simeq f\circ id_X=f$, it follows that $f$ is homotopic to a constant map.

$\left(\Longleftarrow\right)$ Suppose, now, that $f:X\to Y$ is null-homotopic (that is, homotopic to a constant map) for every map $f$ and for arbitrary $Y$. In particular for $Y=X$ and $f=id_X$, it follows that $id_X:X\to X$ is homotopic to a constant map, i.e. that $X$ is contractible.

(b) $\left(\Longrightarrow\right)$ Suppose that $X$ is contractible and that $f:Y\to X$ is a map for $Y$ arbitrary. The argument here is similar to the one in part (a): Define $\widehat{f}=gf:Y\to x_0$ and note that $\widehat{f}$ is necessarily constant and that $h\circ\widehat{f}=h\circ(g\circ f)\simeq id_X\circ f = f$. Hence, $f$ is homotopic to a constant map.

$\left(\Longleftarrow\right)$ Suppose now that $f:Y\to X$ is null-homotopic for every $Y$ and for every $f$. Again, choosing $Y=X$ and $f=id_X$ implies that $id_X:X\to X$ is null-homotopic and hence that $X$ is contractible.   $\square$