If $X$ is a topological space that is not path-connected, does there exist a continuous $f : X \to Y$ such that $f[X]$ is path connected?

I tried to show that such a continuous function didn't exist by doing the following.

Attempt to Prove

Suppose $X$ is not path-connected, then there exist two paths $p, q \in X$ that don't have a path between the,. Let $f : X \to Y$ be a continuous map and assume $f[X]$ is path connected.

Then by definition there exists a continuous $\varphi : [0, 1] \to f[X]$ such that $\varphi(0) = f(p)$ and $\varphi(1) = f(q)\cdots.$

But that's as far as I got, I was thinking of either finding $f^{-1}$ (which may not exist) and then composing to get a path $\gamma : [0, 1] \to X$ and thus arrive at a contradiction, but that would only work if $f$ was a topological embedding (i.e $f$ was a homeomorphism between $X$ and $f[X]$).

I'm also aware of the result that the image of a path-connected space under a continuous map is path-connected, but I don't think I can use that here.

Maybe I'm wrong and there does exist an example where $X$ is a path-connnected space and $f : X \to f[X] \subseteq Y$ is a continuous map with $f[X]$ path-connected, but I can't seem to think up any.

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    $\begingroup$ A constant function? $\endgroup$
    – Randall
    Dec 13 '17 at 15:44
  • $\begingroup$ Let $Y$ be a point... $\endgroup$
    – MooS
    Dec 13 '17 at 15:45

Any constant map $f: X \to Y$ will do this for you (with any space $Y$, no less). It is certainly continuous, and its image path connected. Since constants do this for you, this also means that the stated result is probably useless.

  • $\begingroup$ Thanks for your answer. What if I add the assumption that $f$ is surjective? $\endgroup$ Dec 13 '17 at 15:48
  • 1
    $\begingroup$ Then let $Y$ be a one-point space. $\endgroup$
    – Randall
    Dec 13 '17 at 15:51
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    $\begingroup$ @Perturbative You can even have a continuous bijection! Let $X=[1,2)\cup (2,3)$ and $Y=(0,2)$ and $f(x)=x \bmod 2$. $\endgroup$ Dec 13 '17 at 15:55

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