I could use some help on this question:
Suppose that $Y \subset X$ is a closed, path-connected subset of a topological space $X,$ and suppose that the quotient space $X/Y$ is path connected. Show that $X$ is path connected. Give an example that shows that the conclusion need not hold if we don't assume $Y$ is path connected.
I think I know the general idea, but writing down cleanly seems difficult to me...
Let $q: X \to X/Y$ be the quotient map where $p(x) = [x]$. Let $x_0, x_1 \in X$ and consider $q(x_0), q(x_1) \in X/Y$. There exists a path between $q(x_0) = [x_0]$ and $q(x_1)=[x_1]$ in $X/Y$ since it is path-connected. Call this path $p:[0, 1] \to X$ where $p(0) = [x_0]$ and $p(1)= [x_1]$...
Now I know we need to move this path back up to $X$ maybe by looking at the inverse image through $q^{-1}$. If $p(t) \neq [y]$ for any $t \in [0, 1]$ with $y \in Y$, then I think $q^{-1} \circ p: [0, 1] \to X$ defines the path in $X$ from $x_0$ to $x_1$ (note that $q^{-1}$ itself is not a function but the preimage of $p(t)$ through $q$).
I am a bit shaky on when some path is in $Y$. I think if that's the case, we can just define the path by looking at the preimage of the portion in $X$. Since $Y$ is closed and path-connected, we can take any path in $Y$ and link it continuously with the portion in $X$ (the rigor escapes me here, but I believe we can indeed meet the boundary of $Y$ by a path in $X$. If $Y$ was not closed then I do not think we could necessarily do that).
Is this idea all correct? How would I write this up cleanly?
The statement you are asked to prove is not true. For instance, let $X$ be the topologist's sine curve $$\{(x,\sin(1/x):x\in(0,1]\}\cup\{0\}\times[-1,1]\subset\mathbb{R}^2$$ and $Y=\{0\}\times[-1,1]$. Then $Y$ is closed and path-connected and $X$ is not path-connected. However, the quotient $X/Y$ is path-connected, since the map $f:[0,1]\to X/Y$ given by $f(t)=[(t,\sin(1/t))]$ for $t>0$ and $f(0)=[(0,0)]$ is continuous and surjective.
(The difficulty with trying to take a path in $X/Y$ and lift it up to $X$ as you propose is that the path in $X$ may not approach any single point as it approaches $Y$ from the outside, so there is no way to make it continuous. This is what happens when you try to lift $f$ in the example above: there is no single point of $Y$ you can use for $f(0)$ to make the lift continuous.)