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Here is a variant of Jason's example with a proof that it is not even topologically locally trivial. Let $T$ be a (complex) manifold that admits a morphism $\phi$ onto $\mathbb P^1=\mathbb P^1_{\mathbb C}$ (or $S^2$ if you prefer)prefer) and there exists a point $a\in T$ with $b=\phi(a)\in \mathbb P^1$ such that $\{a\}=\phi^{-1}(b)$. This is satisfied for example if $\phi={\rm id}_{\mathbb P^1}$. Let $\Gamma\subset T\times \mathbb P^1$ be the graph of $\phi$ and let $a\in T$ be a point with $b=\phi(a)\in \mathbb P^1$ and $c\in \mathbb P^1, c\neq b$.

Now let $X=T\times \mathbb P^1\setminus \bigg( \Gamma\cup \big(T\times \{b\}\big)\cup \{(a,c)\}\bigg)$ with the natural projection $\pi:X\to T$. Then every fiber of $\pi$ is isomorphic to $\mathbb P^1\setminus \{0,\infty\}\simeq \mathbb C^*\sim S^2\setminus \{P,Q\}$ (for two points $P,Q\in S^2$).

Claim $\pi$ is not topologically locally trivial near $a\in T$.

Proof Suppose $a$ has a neighbourhood $U\subseteq T$ such that $Y:=\pi^{-1}U\simeq U\times S^2\setminus \{P,Q\}$. Then there exists a projection $p:Y\to S^2\setminus \{P,Q\}$. Consider a circle in $S^2\setminus \{P,Q\}$ that's non-trivial in $H_1(S^2\setminus \{P,Q\}, \mathbb Z)$. Since $p$ is an isomorphism between $\pi^{-1}(a)$ and $S^2\setminus \{P,Q\}$, the same circle lives in $\pi^{-1}(a)$ as well. Then the homology class of the circle can be represented by a "small" circle around the point $(a,c)$ (this point is not in $X$!). Next take a "small" ball inside $T\times \mathbb P^1$ with center at $(a,c)$ that contains the previous "small" circle. By the construction of $X$, the intersection of this ball with $X$ is the entire ball except its center $(a,c)$. Therefore the homology class of that "small" circle in $X$ is trivial. However, this is a contradiction, because it was chosen in a way that its image via $p_*$ would be a nontrivial homology class. $\qquad {\rm Q.E.D.}$

Remark I suppose a similar proof works to show that Jason's example is also not locally trivial.

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Here is a variant of Jason's example with a proof that it is not even topologically locally trivial. Let $T$ be a (complex) manifold that admits a morphism $\phi$ onto $\mathbb P^1=\mathbb P^1_{\mathbb C}$ (or $S^2$ if you prefer). Let $\Gamma\subset T\times \mathbb P^1$ be the graph of $\phi$ and let $a\in T$ be a point with $b=\phi(a)\in \mathbb P^1$ and $c\in \mathbb P^1, c\neq b$.

Now let $X=T\times \mathbb P^1\setminus \bigg( \Gamma\cup \big(T\times \{b\}\big)\cup \{(a,c)\}\bigg)$ with the natural projection $\pi:X\to T$.

Claim $\pi$ is not topologically locally trivial near $a\in T$.

Proof Suppose $a$ has a neighbourhood $U\subseteq T$ such that $Y:=\pi^{-1}U\simeq U\times S^2\setminus \{P,Q\}$. Then there exists a projection $p:Y\to S^2\setminus \{P,Q\}$. Consider a circle in $S^2\setminus \{P,Q\}$ that's non-trivial in $H_1(S^2\setminus \{P,Q\}, \mathbb Z)$. Since $p$ is an isomorphism between $\pi^{-1}(a)$ and $S^2\setminus \{P,Q\}$, the same circle lives on in $\pi^{-1}(a)$ as well. Then the homology class of the circle can be represented by a "small" circle around the point $(a,c)$ (this point is not in $X$!). Next take a "small" ball inside $T\times \mathbb P^1$ with center at $(a,c)$ that contains the previous "small" circle. By the construction of $X$, the intersection of this ball with $X$ is the entire ball except its center $(a,c)$. Therefore the homology class of that "small" circle in $X$ is trivial. However, this is a contradiction, because it was chosen in a way that its image via $p_*$ would be a nontrivial homology class. $\qquad {\rm Q.E.D.}$

Remark I suppose the same a similar proof works to show that Jason's example is similarly also not locally trivial.

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Here is a variant of Jason's example with a proof that it is not even topologically locally trivial. Let $T$ be a (complex) manifold that admits a morphism $\phi$ onto $\mathbb P^1=\mathbb P^1_{\mathbb C}$ (or $S^2$ if you prefer). Let $\Gamma\subset T\times \mathbb P^1$ be the graph of $\phi$ and let $a\in T$ be a point with $b=\phi(a)\in \mathbb P^1$ and $c\in \mathbb P^1, c\neq b$.

Now let $X=T\times \mathbb P^1\setminus \bigg( \Gamma\cup \big(T\times \{b\}\big)\cup \{(a,c)\}\bigg)$ with the natural projection $\pi:X\to T$.

Claim $\pi$ is not topologically locally trivial near $a\in T$.

Proof Suppose $a$ has a neighbourhood $U\subseteq T$ such that $Y:=\pi^{-1}U\simeq U\times S^2\setminus \{P,Q\}$. Then there exists a projection $p:Y\to S^2\setminus \{P,Q\}$. Consider a circle in $S^2\setminus \{P,Q\}$ that's non-trivial in $H_1(S^2\setminus \{P,Q\}, \mathbb Z)$. Since $p$ is an isomorphism between $\pi^{-1}(a)$ and $S^2\setminus \{P,Q\}$, the same circle lives on $\pi^{-1}(a)$ as well. Then the homology class of the circle can be represented by a "small" circle around the point $(a,c)$ (this point is not in $X$!). Next take a "small" ball inside $T\times \mathbb P^1$ with center at $(a,c)$ that contains the previous "small" circle. By the construction of $X$, the intersection of this ball with $X$ is the entire ball except its center $(a,c)$. Therefore the homology class of that "small" circle in $X$ is trivial. However, this is a contradiction, because it was chosen in a way that its image via $p_*$ would be a nontrivial homology class. $\qquad {\rm Q.E.D.}$

Remark I suppose the same proof works to show that Jason's example is similarly not locally trivial.

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