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?

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.

• A constant function? Dec 13 '17 at 15:44
• Let $Y$ be a point...
– 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.
• Thanks for your answer. What if I add the assumption that $f$ is surjective? Dec 13 '17 at 15:48
• Then let $Y$ be a one-point space. Dec 13 '17 at 15:51
• @Perturbative You can even have a continuous bijection! Let $X=[1,2)\cup (2,3)$ and $Y=(0,2)$ and $f(x)=x \bmod 2$. Dec 13 '17 at 15:55