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Let $f,g:X\to Y$ and $Y$ is homeomorphic to $Z$. $Z$ is convex. Then show that $f$ and $g$ are homotopic.

Attempt:

Let $h$ be the homeomorphism between $Y$ and $Z$. Then $h\circ f$ and $h\circ g$ are homotopic. (Because $Z$ is convex.)

Now, we also know that if two maps are homotopic then their composition with a continuous function will also be homotopic.

Thus $h^{-1}\circ h\circ f$ and $h^{-1}\circ h\circ g$ are also homotopic.

hence $f$ and $g$ are homotopic.

Is my proof correct or is there any better solution?

  • 0
    Your proof is correct. – 2017-02-13

1 Answers 1

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Looks good to me. You may want to specify that $$H:X\times I\to Y,\quad H(x,t) = (1-t)f(x)+tg(x)$$ gives the homotopy between $h\circ f$ and $h\circ g$.

For future reference, this is a special case of the following fact:

Any two maps $f,g:X\to Y$ into a contractible space $Y$ are homotopic.

If $Z\subseteq\mathbb{R}^n$ is convex, then for any fixed $z_0\in Z$ the map $$H:Z\times I\to Z,\quad H(z,t) = z(1-t)+tz_0$$ is well-defined and continuous, and therefore gives a homotopy between $id_Z$ and the constant map $c(z_0):Z\to Z$. That is, $Z$ is contractible. Since you have a homeomorphism $h:Y\to Z$, you can form the composition $$H':\quad Y\times I\xrightarrow{\ h\times id_I \ }Z\times I\xrightarrow{\ \ \ }Z\xrightarrow{\ h^{-1} \ }Y$$ which obeys: $$H'(y,0) = h^{-1}(h(y)) = y\quad\text{and}\quad H(y,1) = h^{-1}(z_0)=y_0.$$ So $H'$ gives a homotopy between $id_Y$ and the constant map $c(y_0):Y\to Y$, and thus $Y$ is contractible.

Here's a proof of the above fact. Since $Y$ is contractible, there is a homotopy between the identity and the constant map $id_Y\simeq c(y_0)$. Composing on the right with $f$ and $g$ separately gives $$f\simeq c(y_0)\circ f\quad\text{and}\quad g\simeq c(y_0)\circ g.$$ However, $c(y_0)\circ f = c(y_0)\circ g$, and so it follows that $f\simeq g$.