Homotopic Mappings Relative to a Subset of a Topological Space

# Homotopic Mappings Relative to a Subset of a Topological Space

 Definition: Let $X$ and $Y$ be topological spaces, $A \subset X$, and let $f, g : X \to Y$ be two continuous functions such that $f(a) = g(a)$ for all $a \in A$. Then $f$ is Homotopic to $g$ Relative to $A$ written $f \simeq_A g$ if there is a continuous function $H : X \times I \to Y$ such that: a) $H_0 = f$. b) $H_1 = g$. c) $H_t(a) = f(a) = g(a)$ for all $a \in A$ and for all $t \in I$. If such a function $H$ exists, then $H$ is said to be a Homotopy from $f$ to $g$ relative to $A$.
 Definition: Let $X$ and $Y$ be topological spaces and let $f, g : X \to Y$ be two continuous functions. Then $f$ is Homotopic to $g$ written $f \simeq g$ if there is a continuous function $H : X \times I \to Y$ such that: a) $H_0 = f$. b) $H_1 = g$. If such a function $H$ exists, then $H$ is said to be a Homotopy from $f$ to $g$.

Observe that $f$ being homotopic to $g$ is the same thing as $f$ being homotopic to $g$ relative to $A = \emptyset$.

 Theorem 1: Let $X$ and $Y$ be topological spaces and let $A \subset X$. Then the relation $f$ is homotopic to $g$ relative to $C$ is an equivalence relation on the set of continuous functions $f, g : X \to Y$ such that $f(a) = g(a)$ for all $a \in A$.
• Proof: Let $f, g, h : X \to Y$ be continuous functions such that $f(a) = g(a) = h(a)$ for all $a \in A$.
• Consider the function $H : X \times I \to Y$ defined for all $t \in I$ by:
(1)
\begin{align} \quad H(x, t) = f(x) \end{align}
• Then $H$ is a continuous function since $f$ is a continuous function. Furthermore, $H_0 = f$, $H_1 = f$, and $H_t(a) = f(a) = f(a)$ for all $a \in A$. So $f$ is homotopic to $f$ relative to $A$.
• Suppose that $f$ is homotopic to $g$ relative to $A$. Then there exists a continuous function $H : X \times I \to Y$ such that $H_0 = f$, $H_1 = g$, and $H_t(a) = f(a) = g(a)$ for all $a \in A$ and for all $t \in I$. Define a new function $H' : X \times I \to Y$ by:
(2)
\begin{align} \quad H'(x, t) = H(x, 1 -t) \end{align}
• Then $H'$ is a continuous function since $H$ is a continuous function. Furthermore, $H'_0 = H_1 = g$, $H'_1 = H_0 = f$, and $H'_t(a) = H_{1-t}(a) = f(a) =g(a)$ for all $a \in A$ and all $t \in I$. So $g$ is homotopic to $f$ relative to $A$.
• Let $f$ be homotopic to $g$ relative to $A$ and let $g$ be homotopic to $h$ relative to $A$. Then there exists continuous functions $H' : X \times I \to Y$ and $H'' : X \times I \to Y$ such that $H'_0 = f$, $H'_1 = g$, $H''_0 = g$, $H''_1 = h$, $H'_t(a) = f(a) = g(a)$ for all $a \in A$ and for all $t \in I$, and $H''_t(a) = g(a) = h(a)$ for all $a \in A$ and for all $t \in I$. Define a new function $H : X \times I \to X$ by:
(3)
\begin{align} \quad H(x, t) = \left\{\begin{matrix} H'(x, 2t) & \mathrm{if}\: 0 \leq t \leq \frac{1}{2} \\ H''(x, 2t - 1) & \mathrm{if} \: \frac{1}{2} \leq t \leq 1 \end{matrix}\right. \end{align}
• Then $H$ is a continuous function since $H'$ and $H''$ are continuous and by The Gluing Lemma. Furthermore, [$H_0 = H'_0 = f$, $H_1 = H''_1 = h$, and $H_t(a) = f(a) = h(a)$ for all $a \in A$ and for all $t \in I$. So $f$ is homotopic to $h$ relative to $A$.
• So indeed, the relation of homotopic relative to $A$ is an equivalence relation. $\blacksquare$
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