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.

  • 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 Nov 7 '11 at 11:45

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.

  • $\begingroup$ 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 Nov 7 '11 at 10:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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