# Does an interval and a triangle share the same homotopy type?

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?

• Intervals are contractible; your triangles aren't. – Lord Shark the Unknown Apr 15 '18 at 6:58
• But how to understand it according to Hatcher's introduction? – Sooner Apr 15 '18 at 7:12
• Contracting a side of your triangle does not result in an interval. The result is two intervals identified at their endpoints. – Lord Shark the Unknown Apr 15 '18 at 7:14
• Why? A 0-cell on a triangle is the endpoint of two edges. If you identify this point to another, then two edges should move. – Sooner Apr 15 '18 at 7:22
• You can collapse an edge but you don't produce an interval. Because two other edges are not identified. You now have two points connected with two edges. A flatened circle. – freakish Apr 15 '18 at 7:37

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?

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.

Does this word "collapse" have the same meaning with continuous deformation retract?

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.

So could I think deeply that every simplex could be contractible which means could shrink to a point?

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

$$H:A\times[0,1]\to A$$ $$H(y, t)=xt+(1-t)y$$

• Wow thank you again! So I guess intuitionally( forgive me, I like to seek for the right intuition first), triangle has the same homotopy type with a circle, and if the interior of the triangle is full( that is what I understand "simplex", because all points on the path of two points which belong to the simplex are all in the simplex. So that means it should be like a disk, and I guess that is what you say about star subset), the disk of triangle shape should be contractible. Is that right? – Sooner Apr 15 '18 at 8:44
• I guess it is kind of like the difference between spheres and balls. – Sooner Apr 15 '18 at 8:45
• @Sooner Yes, exactly. Your tringle is sphere. And simplex is ball. Not "like", they are precisely the same up to homeomorphism. Not only homotopy. – freakish Apr 15 '18 at 8:46
• Wow you proposed a problem that I always ignored. What is the difference between homotopy and homeomorphism? I know homotopy means that you need a continuous map between space A and B, and homeomorphism means that we could find continuous bijective maps between A and B. So homeomorphism is the last stage of the homotopy , right? I mean for the $f_t$, the homeomorphism should be the $f_1$ because the $f_1$ is exactly the map we need to find for homeomorphism . – Sooner Apr 15 '18 at 8:54
• Also I remembered that simplex should be concerned about triangles. And the 3-simplex should be a tetrahedron instead of a ball although I know that tetrahedron and ball are the same homotopy type because we can find a way to deform the tetrahedron to a ball continuously. – Sooner Apr 15 '18 at 8:56