Let both $(X , d)$ and $(Y , d)$ are non-trivial connected metric spaces. That is both $X$ and $Y$ contains at least two elements each.

Is it true that there always exist a non-constant continuous function from $(X ,d)$ to $(Y ,d)$. I know that if $(X ,d)$ is connected and $(Y ,d)$ is totally disconnected (i.e connected components of $(Y ,d)$ are the one-point sets) then there cannot exist a non-constant continuous function from $(X , d)$ to $(Y , d)$. Also if $(X ,d)$ is not connected then no matter what is the structure of $(Y ,d)$ there always exist a non-constant continuous function. Furthermore, with the mentioned assumptions if the cardinality of $X$ is less than or equal to $Y$ then is it true that there always exist an injective continuous map from $(X ,d)$ to $(Y ,d)$.

Any help please. Thanks in advance.

  • $\begingroup$ I don’t have time to think about it right now, but it’s true under the stronger assumption that $Y$ is path connected, even if $X$ is Tikhonov rather than metric. Let $x_0,x_1\in X$ with $x\ne y$, and let $f:X\to[0,1]$ be such that $f(x_0)=0$ and $f(x_1)=1$; connectedness of $X$ implies that $f[X]=[0,1]$. Let $y_0,y_1\in Y$ with $y_0\ne y_1$; then there is a path $g:[0,1]\to Y$ such that $g(0)=y_0$ and $g(1)=y_1$. The map $g\circ f$ is a non-constant continuous function from $X$ to $Y$. $\endgroup$ Mar 4 at 8:25
  • $\begingroup$ @ Brian M.Scott Can you please give some hint , why the function f from X to [0 , 1] as you have defined is a continuous function. $\endgroup$
    – m pandey
    Mar 4 at 10:06
  • $\begingroup$ The existence of such a function is an immediate consequence of the the fact that $X$ is a Tikhonov space. $\endgroup$ Mar 4 at 19:22
  • 1
    $\begingroup$ Thanks for the clarification. $\endgroup$
    – m pandey
    Mar 5 at 5:05

Let me address all the points here:

  1. If $X$ is connected and $Y$ totally disconnected, then only constant functions $X\to Y$ are continuous.
  2. If $X$ is disconnected and $Y$ has at least two points then there exists a non-constant continuous function $X\to Y$.
  3. If $X$ and $Y$ are connected then it doesn't mean that there is a non-constant continuous function $X\to Y$. Consider $X=[0,1]$ and let $Y$ be the pseudo-arc. The pseudo-arc is well known to be connected but totally path-disconnected (any such space will work) and so every continuous map $X\to Y$ is constant. This example also answers the question about cardinalities.
  4. If $X$ is any metric space and $Y$ is path connected, both with at least two points, then there is always a non-constant map $X\to Y$. Let $x,y\in X$, $x\neq y$. By Tietze extension theorem the function $\{x,y\}\to[0,1]$ that maps $x$ to $0$ and $y$ to $1$ extends to $A:X\to [0,1]$. We then compose that $A$ with any path $[0,1]\to Y$ connecting two distinct points.
  • $\begingroup$ It is very hard for me to understand the pseudo-arc. However, is it true that if X is connected and Y is path-connected then there exists a non-constant continuous function from X to Y. $\endgroup$
    – m pandey
    Mar 4 at 10:15
  • $\begingroup$ @mpandey I've updated the answer. $\endgroup$
    – freakish
    Mar 4 at 12:39
  • $\begingroup$ As a side note, since $Y$ is not supposed to be compact, there's an example easier than the pseudo-arc: the Knaster–Kuratowski fan $\endgroup$
    – Caffeine
    Mar 4 at 13:25
  • $\begingroup$ @freakish Thank you so much. $\endgroup$
    – m pandey
    Mar 4 at 16:22

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