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If $H$ is a homotopy from $f$ to $g$, and $H'$ is a homotopy of $g$ to $j$, then

$F\left(x,t\right) = \left\{ \begin{array}{lr} H(x,2t) &,\text{ if } 0 \le t \le \frac{1}{2}\\ H'(x,2t-1) &, \text{ if } \frac{1}{2} \le t \le 1 \end{array} \right.\\$

is a homotopy from $f$ to $j$.

I can see that this is a transitive property, but I can't seem to find a way to show that this mapping is continuous.

I know to show continuity I must show that $V \in T' \Rightarrow F^{-1}(V) \in T$, but I can't see how to show that $F^{-1}(V) \in T.$

Anyone have any ideas?

1 Answers 1

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This follows from the pasting lemma. If the domain of $F$ is $X\times[0,1]$, the restriction of $F$ to $X\times[0,1/2]$ is clearly continuous, since it is obtained by composing the continuous functions $H$ and $(x,t)\mapsto (x,2t)$. Similarly, the restriction of $F$ to $X\times[1/2,1]$ is also continuous. Since these sets are closed, by the pasting lemma $F$ is also continuous on their union, which is all of $X\times[0,1]$.

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    What do you mean by composing the continuous functions $H$ and $(x,t) \mapsto (x,2t)$? What's the composition? I'm unfamiliar with the $\mapsto$ notation. – 2017-02-01
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    Define $g:[0,1]\times[0,1]\to[0,1]\times[0,1/2]$ by $g(x,t)=(x,2t)$. Then $g$ is continuous, and $H=F\circ g$. – 2017-02-01
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    Why wouldn't the map $g$ you defined map from $[0,1] \times [0,1] \rightarrow [0,1] \times [0,2]$ since $g(x, 1) = (x,2)$? – 2017-02-01
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    Oops, I meant $g:[0,1]\times[0,1/2]\to[0,1]\times[0,1]$. – 2017-02-01
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    Why did you choose $[0,1]$ for the first factor of the domain and codomain of the products? The initial product space in the homotopy is just an arbitrary space $X$, are you just using $[0,1]$ as an example? – 2017-02-01
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    Oh, sorry, I assumed you were talking about a homotopy of paths. Yeah, the first coordinate should be whatever the domain of your maps are. – 2017-02-01