I am familiar with the following Lifting Theorem:

Let $p: X \rightarrow B$ be a covering with $p(x_0) = b_0$ and let $f: Y \rightarrow B$ be a continuous map with $f(y_0) = b_0$. Assume Y is path-connected and locally path-connected. Then a lift $g$ of $f$ exists (i.e. $g: Y \rightarrow X$, $g(y_0) = x_0$ and $p \circ g = f$) if and only if $$f_*(\pi_1 (Y, y_0)) \subset p_*(\pi_1(X, x_0))$$ and in that case the lift is unique.

I know came across the following slightly different statement:

Let $p: X \rightarrow B$ be a covering with $p(x_0) = b_0$ and let $f: Y \rightarrow B$ be a continuous map with $f(y_0) = b_0$. Assume X is is connected and Y locally path-connected. If $$f_*(\pi_1 (Y, y_0)) \subset p_*(\pi_1(X, x_0))$$ then there exists a unique continuous lift $g: Y \rightarrow X$ of $f$ with $g(y_0) = x_0$ and $p \circ g = f$.

So the assumption that $Y$ is path-connected was replaced by $X$ being connected. Is the existence of the lift then still guaranteed? I know $Y$ being locally path-connected is necessary as can be seen here. But is path-connectedness of $Y$ necessary?

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That will not be enough : indeed the assumption on $\pi_1(Y,y_0)$ only tells us about the path-component of $y_0$, nothing else.

To get a specific counterexample, take $X\to B$ to be the $2$-sheeted connected covering of $S^1$ (so $z\mapsto z^2, S^1\to S^1$), $b_0 = 1, x_0 = 1$; and take $Y = S^1\sqcup \mathbb R$, $y_0 = 0 \in \mathbb R$ and $f: Y\to B$ defined by $id_{S^1}$ on $S^1$ and $x\mapsto \exp(ix)$ on $\mathbb R$

Then clearly, as $\mathbb R$ is contractible, the requirement for $\pi_1(Y,y_0)$ is satisfied, but there is no lift (a lift would yield a section of the $2$-sheeted covering, which does not exist)

So unless we require more than just information about the path-component of $y_0$, path-connectedness of $Y$ is necessary

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  • $\begingroup$ Thanks for the counterexample. I am unfortunately not familiar with sections, is there an elementary way to see that such a lift cannot exist? $\endgroup$ – EinStone Oct 13 '19 at 12:58
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    $\begingroup$ Well since you seem to know about fundamental groups : if there was a lift, restrict it to $S^1\subset Y$, you get $s: S^1\to S^1$ with $p\circ s = id_{S^1}$ where $p:z\mapsto z^2$. In particular $\pi_1(S^1,1) =(id_{S^1})*\pi_1(S^1,1) \subset p_*\pi_1(S^1,1)$, but since $p$ is a $2$-sheeted covering, $p_*\pi_1(S^1,1)$ has index $2$ in $\pi_1(S^1,1)$, so that's a contradiction $\endgroup$ – Maxime Ramzi Oct 13 '19 at 13:02

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