path connectedness of a preimage Suppose $p:X \rightarrow Y$ is a fibration such that $Y$ is path connected and $p^{-1}\{y\}$ is path connected for some $y \in Y$. Could anyone please show me that with all these conditions that $X$ is also path-connected. Thank you all for helping

I will use more common notation. So let $\pi:E\to B$ be a fibration with $B$ path connected and $b\in B$ be such that $\pi^{-1}(b)$ is path connected.
Pick $x, y\in E$ and consider $\lambda:I\to B$ a path such that $\lambda(0)=\pi(x)$ and $\lambda(1)=\pi(y)$. Such path exists because $B$ is path connected. Let $\{*\}$ be a space with exactly one point $*$ and put
$$f:\{*\}\times I\to B$$
$$f(*, t)=\lambda(t)$$
We can apply the homotopy lifting property to this map, because we have a constant map $g:\{*\}\to E$, $g(*)=\lambda(0)$ which is equal to $g(*)=f(*,0)$ and so $\pi\circ g = f\circ i$ where $i$ is the embeding of $\{*\}\to\{*\}\times\{0\}$. In other words we have a commuting diagram
$$\require{AMScd}
\begin{CD}
\{*\} @>g>> E\\
@VViV     @VV\pi V     \\
\{*\}\times I @>f>> B
\end{CD}
$$
Therefore there exists
$$F:\{*\}\times I\to E$$
such that $\pi\circ F=f$. Note that $F(*,0)\in\pi^{-1}(x)$ and $F(*, 1)\in\pi^{-1}(y)$. Therefore this map induces a path between some point in $\pi^{-1}(x)$ and some point in $\pi^{-1}(y)$, namely $t\mapsto F(*, t)$. So if we knew that every fiber is path connected then we are done because we already know how to connect different fibers.
But if $B$ is path connected then every two fibers are homotopy equivalent. And since homotopy equivalence preserves (path) connectness then every fiber is path connected because $\pi^{-1}(b)$ is. $\Box$
