# Assumptions in Lifting Theorem

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?

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
• 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 – Maxime Ramzi Oct 13 '19 at 13:02