# If $f:X\rightarrow Y$ has the path lifting property the so does $f|_A$?

Let $f:X \rightarrow Y$ be a continuous surjective map between topological spaces that has the path lifting property. This means that given a path $\alpha : I \rightarrow Y$ in $Y$ and a point $x_0 \in f^{-1}(\alpha(0))$ there exists a path $\tilde{\alpha}:I\rightarrow X$ such that $f\circ \tilde{\alpha}=\alpha$, $\tilde{\alpha}(0) = x_0$. Let $A\subseteq X$. Then I have the following question -

Does $f|_A:A\rightarrow f(A)$ have the path lifting property?

If I have a path $\beta :I \rightarrow A$ and a point $a\in f^{-1}(\beta(0))\cap A$ then I can lift it to a path $\tilde{\beta}$ in $X$. But I don't see why this should be a path in $A$, that is why $\tilde{\beta}(I)\subseteq A$? So my guess is that it false but I can't come up with a counter example.

Thank you.

Here's a counterexample for which $A$ is closed in $X$. Let $X=\mathbb{R}^2$, $Y=\mathbb{R}$, and $f(x,y)=x$. This has the path lifting property (to lift a path from $Y$ to $X$, just make the second coordinate be constant). But for $A=[0,1]\times\{0\}\cup[1,2]\times\{1\}$, the restriction $f|_A:A\to [0,2]$ does not have the path lifting property, since you cannot lift a path from $0$ to $2$ to $A$.
You can also easily modify this example to have $A$ be open in $X$ (just take an open ball around the $A$ above small enough so that it is still disconnected).
Consider the covering map $p:\mathbb{R}\to S^1,\quad t\mapsto e^{it}$, and set $A=[0,2\pi)\subset\mathbb{R}$. Then $p(A)=S^1$, and the restriction $p|_A$ does not have the path lifting property.