0
$\begingroup$

Let $X$ and $Y$ be topological spaces. Assume $Y$ is contractible (hence, path- connected).

Let $f,g: X \to Y$ be continuous maps. At any fixed $x\in X$, there is a path $P_x: [0,1]\to Y$ from $f(x)$ to $g(x)\in Y$ such that $P_x(0)=f(x)$ and $P_x(1)=g(x)$

We define $F: X\times [0,1]\to Y$ by $F(x,t) = P_x(t)$. Now, $$F(x,0)= P_x(0)=f(x), F(x,1)= P_x(1)=g(x)$$ for any $x\in X$ (This is why $F$ is homotopy-like). Clearly, $F$ is continuous at time $0$ and at time $1$. Is $F$ necessarily continuous at any time $t$?

Thanks in advance for any help.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ In your example you make merely a funtion $F$ such that for any $t$ the function $F_t: x \mapsto F(x, t)$ is contnuous, of course this is too weak for have the (global) continuity of $F$. Anyway if you parametrize the paths by the homotopy contraction maps you get a (continuous) homotopy $F$ from $f$ to $g$. Let $G: X \times I\to Y: 1_Y\ \tilde\ \ C(y_0)$ (contraction homotopy from identity map to costant map on the point $y_0$). then join the maps $F(x, t):= G(f(x), 2t)\ t\in[0,1/2]$ and $F(x, t):= G(g(x), 2-2t)\ t\in[1/2,1]$ . $\endgroup$
    – Buschi Sergio
    Commented Nov 7, 2011 at 11:45

1 Answer 1

Reset to default
3
$\begingroup$

This proof cannot work since then every two maps $X \to Y$, where $Y$ is path-connected, are homotopic - which is false. On the other hand, if $Y$ is contractible, then every map $X \to Y$ is homotopic to a constant map (since this true for the identity $X \to X$), thus every two maps $X \to Y$ are homotopic.

Even if $Y$ is contractible, in your proof the paths $P_x$ may be choosen so arbitrarily that $F$ is not continuous. Indeed, there are maps $F : [0,1] \times [0,1] \to \mathbb{R}$ such that $F(-,0)$, $F(-,1)$ and all $F(t,-)$ are continuous, but $F$ is not.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Oh, I see. Thanks Martin. I already knew the other proof you described in the last line. But I was stuck why I could not progress the way I described in my question. $\endgroup$
    – Chulumba
    Commented Nov 7, 2011 at 10:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .