## 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

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

$\tag*{\blacksquare}$