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?
Best Answer
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