Main Idea: Nothing to it. Homotopy:
\(A \homotopic \Delta \homotopic S^1\)
\(a \homotopic d \homotopic o \homotopic S^1\)
\(B \homotopic 8 \homotopic S^1 \vee S^1\)
\(b \homotopic o \homotopic S^1\)
\(C \homotopic *\)
\(c \homotopic l \homotopic *\)
\(D \homotopic S^1\)
\(d \homotopic o \homotopic S^1\)
\(E \homotopic *\)
\(e \homotopic d \homotopic S^1\)
\(F \homotopic *\)
\(f \homotopic *\)
\(G \homotopic *\)
\(g \homotopic 8 \homotopic S^1 \vee S^1\)
\(H \homotopic *\)
\(h \homotopic l \homotopic *\)
\(I \homotopic *\)
\(i \homotopic \theset{*_1, *_2}\)
\(J \homotopic *\)
\(j \homotopic i \homotopic \theset{*_1, *_2}\)
\(K \homotopic *\)
- \(k \homotopic K \homotopic *\)
\(L \homotopic *\)
- \(l \homotopic *\)
\(M \homotopic *\)
- \(m \homotopic *\)
\(N \homotopic *\)
- \(n \homotopic *\)
\(O \homotopic S^1\)
- \(o \homotopic S^1\)
\(P \homotopic D \homotopic S^1\)
- \(p \homotopic P \homotopic S^1\)
\(Q \homotopic O \homotopic S^1\)
- \(q \homotopic p \homotopic o \homotopic S^1\)
\(R \homotopic D \homotopic S^1\).
- \(r \homotopic l \homotopic S^1\)
\(S \homotopic *\)
- \(s \homotopic S \homotopic *\)
\(T \homotopic *\)
- \(t \homotopic l \homotopic *\)
\(U \homotopic *\)
- \(u \homotopic U \homotopic *\)
\(V \homotopic *\)
- \(v \homotopic V \homotopic *\)
\(W \homotopic *\)
- \(w \homotopic W \homotopic *\)
\(X \homotopic *\)
- \(x \homotopic X \homotopic *\)
\(Y \homotopic *\)
- \(y \homotopic v \homotopic *\)
\(Z \homotopic *\)
- \(z \homotopic Z \homotopic *\)
This results in a partition of the alphabet into the following homotopy types:
\(\theset{A, D, O, P, Q, R, S^1}\) \(\union \theset{a,b,d,e,g,o,p,q}\)
\(\theset{C, E, F, G, H, I, J, K, L, M, N, S, T, U, V, W, X ,Y, Z, *}\) \(\union \theset{c,f,h,k,l,m,n,r,s,t,u,v,w,x,y,z}\)
\(\theset{B, S^1 \vee S^1}\)
\(\theset{i, j, \theset{*, *}}\)
Homeomorphisms: ignore ligatures!!
\(\theset{A, R}\) Can remove a point to obtain two components homeomorphic to \(\theset{I, F}\) respectively.
\(\theset{D, O, S^1}\) These all have no single point that can be removed to disconnect the space.
\(\theset{B,S^1 \vee S^1}\) Remove point at junction
\(\theset{C, G, I, J, L, M, N, S, U, V, W, Z, [0,1]}\) These all have a point that can be removed to yield two components, but no points that yield three. (Intuitively, all can be obtained by twisting a straight wire.)
\(\theset{E, F, T, Y, \bigvee_{i=1}^3 [0,1]}\) These all have a point that can be removed to yield 3 connected components homeomorphic to \(I\). This is the “pasting” point in the vee.
\(\theset{H, K, \bigvee_{i=1}^5 [0,1]}\) Can remove two points to disconnect each into five components.
\(\theset{P, Q, S^1 \vee [0,1]}\) Both contain a nontrivial loop.
\(\theset{X, \bigvee_{i=1}^4 [0,1]}\) Can remove one point to separate into four components.
Solution
The classifications depend on the chosen font or drawing; for example, a lowercase letter can have one or two components.