# Understanding homotopy equivalence in a special case

I'm beginning to learn algebraic topology from Hatcher, and I'm trying to get an intuitive grasp on homotopy equivalence. Hatcher says these two shapes are homotopy equivalent: Let's call the left one A and the right one B. I can easily see visually how to "stretch" A to get B, but I'm having trouble understanding how to apply the actual, formal definition of homotopy equivalence.

Specifically, we need continuous functions $$f : A\to B$$ and $$g : B\to A$$ such that $$fg$$ is homotopic to $$id_B$$ and $$gf$$ is homotopic to $$id_A$$. But I can't really see how to do this. For $$g$$, we can map the whole vertical bar of $$B$$ (including the endpoints) to the point at the center of $$A$$, and the other parts of $$B$$ besides the vertical bar to the round parts of $$A$$ in the obvious way. This is continuous. But what can we do for a continuous $$f$$, such that the compositions $$fg$$ and $$gf$$ are homotopic to the identity maps?

• Embed them in $\Bbb R^2$; the former is $\Bbb S^1\lor\Bbb S^1$, and the latter is a circle with a diameter, and then contracts this diameter. Jan 14 at 7:18
• There is another way if $X$ is a CW-complex and $Y$ be a contractible subcomplex of $X$, then $\frac{X}{Y}$ is homotopically equivalent to $X$. Take $X=B$ and $Y=$ the diameter of $B$. Note that $B\simeq\frac{B}{\text{diameter}}=A$. Notice that both $A,B$ are graphs i.e. one-dimensional CW-complex. Jan 14 at 7:29
• $f$ may map the center of $A$ to the center of the vertical bar of $B$, and also glue together some parts of the left and right loops of $A$ to make the vertical bar of $B$. Jan 14 at 7:47
• I think I understand now. Thank you! Jan 14 at 7:51

Let $$S(p)$$ denote the plane circle with radius $$1$$ and center $$p \in \mathbb R^2$$ and $$D = \{0\} \times [-1,1]$$. Let $$S_l(p)$$ and $$S_r(p)$$ denoet the left and right closed halfcircle of $$S(p)$$. Let $$p^- = (-1,0), p^+ = (1,0)$$ and $$p^0 = (0,0)$$.
Then $$A = S(p^-) \cup S(p^+) , B = S(p^0) \cup D .$$ Define $$f : A \to B$$ by projecting $$S_r(p^-)$$ and $$S_l(p^+)$$ horizontally to $$D$$ and by translating $$S_l(p^-)$$ horizontally to $$S_l(p^0)$$ and translating $$S_r(p^+)$$ horizontally to $$S_r(p^0)$$.
Define $$g : B \to A$$ by contracting $$D$$ to $$p^0$$ and mapping $$S_l(p^0)$$ in the obvious way to $$S(p^-)$$ and $$S_r(p^0)$$ to $$S(p^+)$$.
The explicit construction of homotopies $$gf \simeq id_A$$ and $$fg \simeq id_B$$ is somewhat tedious, but I am sure you can see how that works. The essence is to understand that the identity $$id_S$$ on a circle $$S$$ is homotopic to map $$\phi : S \to S$$ contracting a closed half circle $$H \subset S$$ to a point $$h \in H$$. This homotopy keeps $$h$$ fixed and "stretches" the open halfcircle $$S \setminus H$$ to $$S \setminus \{h\}$$.