## Problems

August 2020

#### Problem 10

*Show that a space \(X\) is contractible if and only if every map \(f: X \rightarrow Y\), for arbitrary \(Y\), is nullhomotopic. Similarly, show \(X\) is contractible if and only if
every map \(f: Y \rightarrow X\) is nullhomotopic.*

If \(X\) is contractible, then its identity map
\(\mathbb{I}: X \rightarrow X\) is nullhomotopic. Let \(h: X \times I \rightarrow X\) be a homotopy between \(\mathbb{I}\) and a constant map
(This map exists since the identity is nullhomotopic). Then the composite \(fh: X \times I \rightarrow Y\) is a homotopy from \(f\) to a constant map. Conversely,
suppose *every* map \(f: X \rightarrow Y\) for arbitrary \(Y\) is nullhomotopic. Then the identity map falls under this category (we can just set \(Y=X\)), and \(X\) is
therefore contractible.

Next, we show that \(X\) is contractible if and only if every map \(f: Y \rightarrow X\) is nullhomotopic. Suppose \(X\) is contractible. Then we can define a homotopy
\(h: X \times I \rightarrow X\) between the the identity and a constant map. It follows that the composite \(hf\) is a homotopy between \(f\) and a constant map (\(h\) is continuous,
\(h(f(Y), 0) = f(Y)\) and \(h(f(Y), 1) = c_0\) for some point \(c_0 \in X\). Conversely, if every map \(f: Y \rightarrow X\) is nullhomotopic, then, setting \(Y=X\), the identity map
must also be homotopic.

I suspect I didn't use the proper notation here for composing functions where the domains of some functions are products, but the logic still holds (I hope!).
\(\tag*{$\blacksquare$}\)

### Problem 11

*Show that \(f: X \to Y\) is a homotopy equivalence if there exist maps \(g, h: Y \to X\) such that \(fg \simeq \mathbb{I}\) and \(hf \simeq \mathbb{I}\). More generally, show that \(f\) is a homotopy equivalence if \(fg\) and \(hf\) are homotopy equivalences.*

\(f\) is a homotopy equivalence if there exists some map \(j: Y \to X\) such that \(fj = \mathbb{I}_Y\) and \(jf = \mathbb{I}_X\). We can construct \(j\) by composing the given maps: \(j = hfg: Y \to X\). Then we have

\[f(hfg) = f(hf)g \simeq f \mathbb{I}_X g = fg \simeq \mathbb{I}_Y\]

and

\[(hfg)f = h(fg)f \simeq h \mathbb{I} f \simeq hf \simeq \mathbb{I}_X\]

Now the problem asks us to prove the same thing by assuming instead that there exist maps \(g, h: Y \to X\) such that \(fg\) and \(hf\) are homotopy equivalences. If \(fg\) is a homotopy equivalence, then there exists a map \(k: Y \to Y\) such that \(kfg \simeq \mathbb{I}_Y\) and \(fgk \simeq \mathbb {I}_Y\). A similar map \(d: X \to X\) must exist for \(hf\).

\(f\) is a homotopy equivalence if there exists some map \(j: Y \to X\) such that \(fj \simeq \mathbb{I}_Y\) and \(jf \simeq \mathbb{I}_X\). We can form \(j\) by composing the above functions like so: \(j = dhfgk: Y \to X\). Then we have

\[ f (dhfgk) = f(dhf)gk \simeq f \mathbb{I}_X gk = fgk \simeq \mathbb{I}_Y \]

and

\[ (dhfgk) f = dh(fgk)f \simeq dh \mathbb{I}_Y f = dhf \simeq \mathbb{I}_X \]

\[\tag*{$\blacksquare$}\]