According to Hatcher's book, Chapter 0 page 12. He introduces one way to find homotopy equivalence instead of thinking deformation retract. He said we could collapse subspace to a point. So I have three questions.
1. I guess if we have a triangle, which is composed of three 0-cell and 3 one-cell, we shrink one edge, or collapse one edge to a point so that we get an interval, so they share the same homotopy type?
2. Does this word "collapse" have the same meaning with continuous deformation retract?
3. So could I think deeply that every simplex could be contractible which means could shrink to a point?
2026-03-29 20:45:46.1774817146
Does an interval and a triangle share the same homotopy type?
171 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
1
You don't get an interval. You get two points connected by two edges. That's because you didn't identify two other edges. Things don't "move" automagically. I know it is helpful sometimes to visualize it like that but the truth is that no such thing happens. You didn't identify them, so they are distinct.
In your construction the homotopy type doesn't change and in fact it is the homotopy type of a circle. Interval on the other hand has the homotopy type of a point. These two are not the same.
If $A\subseteq X$ then collapsing $A$ means creating the space $X/A$ which is the quotient space with relation given by $x\sim y$ if and only if $x=y$ or $x,y\in A$. Collapsing a subspace may change homotopy type (e.g. take any noncontractible space and collapse everything). In some cases it does not, e.g. when $(X,A)$ is a CW pair and $A$ is a contractible subcomplex then the quotient map $X\to X/A$ is a homotopy equivalence (see: Hatcher). This is exactly the case of your triangle.
Yes, every simplex is contractible. Or more generally: every star subset of a normed vector space is contractible. Star subset is a subset $A\subseteq V$ such that there is a point $x\in A$ with $$\forall_{y\in A}\ \ \forall_{t\in[0,1]}\ \ tx+(1-t)y\in A$$
So $A$ is contractible via
$$H:A\times[0,1]\to A$$ $$H(y, t)=xt+(1-t)y$$