Topology and Groups

Two continuous maps $f,g:X → Y$ are homotopic if there exists a continuous map $H: X × I → Y$ such that $H(x, 0) = f(x)$ and $H(x, 1) = g(x)$ for all $x ∈ X$.
A pair of spaces $X,Y$ are homotopy equivalent if there exist continuous maps $f: X → Y$ and $g: Y → X$ such that $g ∘ f$ is homotopic to $\text{id}_X$ and $f ∘ g$ is homotopic to $\text{id}_Y$.
The fundamental group of $X$ based at $b$ is the set of homotopy classes of loops based at $b$, denoted $π_1(X, b)$. The group operation is concatenation of loops, with the identity element being the homotopy class of the constant loop at $b$.
Let $[ℓ]$ be the homotopy class of a loop $ℓ$ based at $b$. Then $[ℓ]^{-1} = [ℓ^{-1}]$ where $ℓ^{-1}$ is the loop traversed in the opposite direction. This is well-defined since the homotopy class of $ℓ^{-1}$ is independent of the choice of representative of $[ℓ]$.

Define what it means for a space to be contractible. Prove that $ℝ^n$ is contractible but $ℝ^2 ∖\{0\}$ is not. [You may assume facts about $π_1(𝕊^1)$ that you state clearly.]

A space is contractible if it is homotopy equivalent to a point. $ℝ^n$ is contractible since the homotopy $H: ℝ^n × I → ℝ^n$ defined by $H(x, t) = (1-t)x$ is a homotopy between the identity map and the constant map at 0. $ℝ^2 ∖\{0\}$ is not contractible since $ℝ^2 ∖\{0\}$ is homotopy equivalent to $𝕊^1$ and hence has fundamental group $ℤ$.

Define the terms abstract simplicial complex and triangulation of a space. [You do not need to define the topological realisation of a simplicial complex.]
Explain why a triangulation of $𝕊^2$ must have at least 4 vertices.
State the Simplicial Approximation Theorem, explaining the terms simplicial map and subdivision.
Let $X$ and $Y$ be topological spaces. Assume that $X$ is homotopy equivalent to $𝕊^2$ and that $Y$ is homotopy equivalent to $𝕊^3$. Prove that every continuous map $X → Y$ is homotopic to a constant map.
[You may assume that the $n$-sphere $𝕊^n$ can be triangulated and that $𝕊^n$ minus a point is homeomorphic to $ℝ^n$, but you may not assume that $X$ and $Y$ can be triangulated.]

An abstract simplicial complex is a pair $(V, Σ)$, where $V$ is a set (called vertices) and $Σ$ is a set of non-empty finite subsets of $V$ (called simplices) such that (i) for each $v ∈ V$, the 1-element set $\{v\}$ is in $Σ$; (ii) if $σ$ is an element of $Σ$, so is any non-empty subset of $σ$.
A triangulation of a space $X$ is a simplicial complex $K$ together with a choice of homeomorphism $|K| → X$.
Start with a $2$-simplex $σ$. Any edge $e$ of $σ$ must belong to another $2$-simplex $τ_e$. But that means $τ_e≠τ_{e'}$ when $e'$ is another edge of $σ$, since otherwise $τ$ would have all vertices in common with $σ$. So we need at least three further $2$-simplices. In other words, no closed surface can be triangulated with less than four triangles.
The Simplicial Approximation Theorem states that every continuous map $f: X → Y$ from a space $X$ to a space $Y$ can be approximated by a simplicial map $g: K → L$ from a simplicial complex $K$ to a simplicial complex $L$. A simplicial map is a map that sends each simplex in $K$ to a simplex in $L$.
A subdivision of a simplicial complex $K$ is a simplicial complex $K'$ together with a homeomorphism $h: |K'| → |K|$ such that, for any simplex $σ'$ of $K'$, $h(σ')$ lies entirely in a simplex of $|K|$ and the restriction of $h$ to $σ'$ is affine.
It is enough to prove every continuous map $𝕊^2 → 𝕊^3$ is homotopic to a constant map (sheet 1 Q9). By Simplicial Approximation theorem, every map $𝕊^2 → 𝕊^3$ is homotopic to a simplicial map $f$. Take a point $x$ in the interior of $𝕊^3$ then $x$ is not in the image of $f$. But $𝕊^3-\{x\}$ is homeomorphic to $ℝ^3$, so it is contractible, so $f$ is null-homotopic. Since $𝕊^3$ is path-connected, all constant maps are homotopic.