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Recall an object in an extensive category is connected if, given a non-trivial coproduct decomposition of it precisely one summand is initial. This is also equivalent to its covariant hom functor preserves coproducts.

Is the following claim true?

Claim. For a continuous map $f:X\to Y$ with $Y$ connected, TFAE.

  • It's a connected object of $\mathsf{Top}_{/Y}$;
  • Its fibers are connected.

I think I see that the former implies the latter, but I am having trouble with the converse. Given $f\to \alpha\amalg \beta$ in $\mathsf{Top}_{/Y}$ then the connectedness of the fibers gives fiberwise factorizations of the bundle map, either through fibers of $\alpha$ or $\beta$. However, I need all these factorization to go through exactly one of $\alpha,\beta$ and I am not sure how to prove this happens. Perhaps this is where the connectedness of $Y$ comes in, but I am not sure how to proceed.

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This isn't true in either direction. For instance, consider the squaring map $f:S^1\to S^1$. This is connected as an object over $S^1$, since the total space is connected. But its fibers are disconnected (each fiber is two points).

In the other direction, consider the obvious map $f:\{0\}\coprod (0,1]\to [0,1]$. Each fiber is a single point, but this is disconnected as a space over $[0,1]$.

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  • $\begingroup$ Thanks for the examples. I was careless and expected connected in slices to be fiberwise connectedness. $\endgroup$ – Arrow Dec 10 '17 at 23:34

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