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We say that paths $\alpha, \beta: I \to X$ with common initial point $\alpha(0)=\beta(0)$ and common terminal point $\alpha(1)=\beta(1)$ are homotopic provided that there is a continuous function $H:I \times I \to X$ such that $H(t,0)=\alpha(t)$ and $H(t,1)=\beta(t)$ for $t \in I$ and $H(0,s)=\alpha(0)=\beta(0)$ and $H(1,s)=\alpha(1)=\beta(1)$ for $s \in I$.

Since a path is a continuous function, I would expect that each map between two homotopic paths, $\alpha$ and $\beta$, is a path, and therefore continuous.

My questions are: is the above true, is it guaranteed by continuity of $H$, and if so, how do we prove it?

On a related note, intuitively we pass from $\alpha$ to $\beta$ as t varies over $[0,1]$. In general, can we explicitly realize a map 'in between' $\alpha$ and $\beta$ by fixing a $t$?

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2 Answers 2

up vote 2 down vote accepted

Of course.

The inclusion map $I \times \{t\} \rightarrow I \times I$ is continuous, therefore the composite $$ I \times \{t\} {\rightarrow} I \times I \stackrel{H}{\rightarrow} X $$

is continuous. The assumption on endpoints shows this is a path from $\alpha(0)$ to $\alpha(1)$.

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2  
It is good to think of a homotopy $h:a \simeq b$ as a "deformation of $a$ into $b$ through the intermediate paths". –  Ronnie Brown Feb 10 '13 at 15:41
    
I wasn't expecting it to be that simple! Thanks. –  Alex Petzke Feb 12 '13 at 5:12

It is in fact guaranteed by the continuity of $H$. To see this, let's fix an $s\in I$, and define a map $i_s\colon I\to I$ by $i_s(t)=(t,s)$. This is clearly continuous. Then the "in between" path $h_s:=H(-,s)$ is just the composition of the continuous functions $i_s$ and $H$. Hence, $h_s$ is itself continuous.

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