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Let $A\subset\mathbb R^n$ be star-convex, that is there exists $a_0\in A$ such that for each $a\in A$, the line segment from $a_0$ to $a$ is contained in $A$. Define $H:A\times[0,1]\to\mathbb R^n$ by $H(a,t)=(1-t)a + ta_0$.

The book (Dieck's Algebraic Topology) claims that $H$ is a null homotopy of the identity, and hence star-convex sets are contractible.

I see that a null homotopy of a map $f$ is a homotopy between $f$ and a constant map, and that a null homotopy of the identity is a contraction of the underlying space. However, I do not see how the homotopy $H$ defined here is a null homotopy, or that it is a null homotopy of the identity. Can anyone help me to understand this?

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Note that $H(a,t)$ is obviously continuous: addition and scalar multiplication of continuous functions $\mathbb R^n\to \mathbb R^n$ yields new continuous functions from old. For each $a \in \mathbb R^n$, $H(a,0) = (1-0)a + 0\cdot a_0 = a$. On the other hand, $H(a,1) = (1-1)a + 1\cdot a_0 = a_0$. Therefore $H(\cdot,0)$ is the identity map, while $H(\cdot, 1)$ is a constant map at $a_0$. This demonstrates that the map is a contraction.

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  • $\begingroup$ Where does the "null homotopy" come into play exactly? The fact that $H(\cdot,0)$ is the identity map? Would I be correct in saying that $H$ is a homotopy between the identity map and the constant map $a_0\mapsto a_0$? I am just beginning to study these concepts and it is very confusing to me. $\endgroup$
    – Math1000
    Dec 7, 2019 at 1:32
  • $\begingroup$ Note that the constant map is emphatically not just $a_0 \mapsto a_0$ but $a \mapsto a_0$ for all $a \in A$. The word “null” in null homotopy comes from thinking of constant maps as somehow “zero”. It is not more complicated than that. $\endgroup$
    – Rylee Lyman
    Dec 7, 2019 at 1:35
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    $\begingroup$ It's just terminology: A function $f$ is null-homotopic if it is homotopic to a constant function. The homotopy itself is what is called the null-homotopy. $\endgroup$
    – Matematleta
    Dec 7, 2019 at 1:38
  • $\begingroup$ @RyleeLyman Yes, $a\mapsto a_0$ for all $a\in A$ is what I meant. And I can see how "null" would be associated with constant maps. Thank you. $\endgroup$
    – Math1000
    Dec 7, 2019 at 1:40
  • $\begingroup$ To make sure I understand though: $H$ is a homotopy between the identity map on $A$ and the constant map $a\mapsto a_0$ on $A$, correct? $\endgroup$
    – Math1000
    Dec 7, 2019 at 1:43

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