Is restriction of a chart is a chart necessarey in the Definition of Fibre Bundle

Question

This is the definition of Fibre Bundle from the notes James F Davis and Paul Kirk:

I think the condition 3 is superfluous, because if you have a chart over $U$, then $\phi : U \times F \rightarrow p^{-1}(U)$ such that $\phi\circ p= p_U$. The restriction of this map to any open set $V\subset U$ automatically gives a map $\phi |_V : V\times F \rightarrow p^{-1}(V)$ such that $$\phi|_{V}\circ p= p_V$$

I do not understand why we need 3 as a condition to be a fibre bundle as condition 3 automatically follows from condition 1.

I have seen the same definition in Bott and Tu also. But in Norman Steenrod's book "Topology of Fiber Bundle" this condition is not required. I guess I am missing out something.

Answer 1

Note that an atlas for a manifold (and hence a fibre bundle) in general does not contain all possible charts. So the statement:

For each chart in our atlas condition (1) must be satisfied (which is actually the definition of being a chart respecting the local trivialisation condition).

does not imply the statement:

Every object which satisfy condition (1) is a chart in our atlas.

In particular, the second statement is false in the context of defining atlases, since there can be incompatible charts.

So your justification is wrong: while you have shown that restrictions of charts to open subsets are charts, they may or may not be a chart that we want.

On the other hand, (3) is implied by (4) and (5) together. To see this, one notes that

1. As you wrote in the question, the restriction of a chart is a chart.
2. It remains to check whether this chart is compatible (condition (4)) with other charts. But we know that firstly it is compatible with its "parent" with transition map being the identity. From this you can conclude that it is compatible with every other chart that its parent is compatible with.
3. Since an atlas is maximal by your definition, it must include this guy which is compatible with all other guys.

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In conclusion: (3) is superfluous, but not for the reason you gave.

On the other hand, sometimes for exposition it can be more convenient to include superfluous conditions as part of the definition rather than having to prove a separate lemma asserting the truth of the condition. I imagine in the books where (3) is stated explicitly as a condition, in subsequent proofs they frequently refer to this condition in various constructions.