Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm having a little trouble understanding several topics from algebraic topology. This question covers a range of topics I have been looking at.

Can anyone help? Thanks!

Suppose $X$ and $Y$ are connected manifolds, $X$ is simply connected, and the universal cover of $Y$ is contractible. Why is every continuous mapping from $X$ to $Y$ homotopic to a constant?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

First observe that if $g: X \to Y$ is a map with $Y$ contractible then $g$ is homotopic to a constant map. For let $h_t: Y\to Y$ be such that $h_t(y) = y, h_1(y) = y_0$. Then $g\circ h_t$ is a homotopy from $g$ to the constant map $y \mapsto g(y_0)$. Similarly, if we precompose a map that is homotopic to a constant with another map, then the composition is also homotopic to a constant.

Now let $p:\tilde Y \to Y$ be a cover of $Y$. A basic fact about covering spaces is that a map $f: X \to Y$ lifts to a map $\tilde f : X \to \tilde Y$ if $f_* \pi_1(X) \subset p_* \pi_1 (\tilde Y)$. So if $X$ is simply connected, this is always true. So in your case, taking $\tilde Y$ to be the universal cover, any map $f: X\to Y$ lifts to a map $\tilde f: X \to \tilde Y$ and $\tilde f$ must be homotopic to a constant map since $\tilde Y$ is contractible. Therefore $f = p \circ \tilde f$ is also homotopic to a constant map.

share|improve this answer

Let $\tilde{Y} \xrightarrow{\pi} Y$ be the universal cover of $Y$. Since $X$ is simply connected, any continuous map $X \xrightarrow{f} Y$ can be factorized as a continuous map $X \xrightarrow{\tilde{f}} \tilde{Y} \xrightarrow{\pi} Y$. Since $\tilde{Y}$ is contractible, there is a point $y \in \tilde{Y}$ and an homotopy $h$ between the identity map on $\tilde{Y}$ and the constant map $y$ :

$h : \begin{array}{c}\tilde{Y} \xrightarrow{id} \tilde{Y} \\ \Downarrow \\ \tilde{Y} \xrightarrow{y} \{y\}\end{array}$

Composing this homotopy with $\tilde{f}$ and $\pi$, you get an homothopy $h'(t,x) = \pi(h(t,\tilde{f}(x))$

$h': \begin{array}{rcl}X \xrightarrow{\tilde{f}} & \tilde{Y} \xrightarrow{id} \tilde{Y} &\xrightarrow{\pi} Y \\ &\Downarrow &\\ X \xrightarrow{\tilde{f}} & \tilde{Y} \xrightarrow{y} \{y\} & \xrightarrow{\pi} \{\pi(y)\} \end{array}$ between $f$ and the constant map $\pi(y)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.