Suppose f:A→B and g:B→A are both surjective, does this imply that there is a bijection between A and B.

I was told that the statement above is true only with Axiom of Choice. Can someone provide an example of why that is the case?

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# Tag: surjection

## Proving cardinality given surjection from A to B and B to A

## analysis – Show that the two-function product surjection is surjection.

## Surjection of a space Frechet involves Frechet?

## combinatory – Using the number of surjection

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Suppose f:A→B and g:B→A are both surjective, does this imply that there is a bijection between A and B.

I was told that the statement above is true only with Axiom of Choice. Can someone provide an example of why that is the case?

Show that the two-function product surjection is surjection.

Solution. (What I tried) Be $g$ and $h$ surjection functions. If $g: A rightarrow B$ is surjection then for all $y in B$ there is x in A such that y=g(x). Similarly, if h: B C is overjetora then for all z in C there is w in B such that z=h(w). But I need to show that my function k(x)=g(x)cdot h(x) is overjetora. Can someone help me please?

Let $ E $ to be a separable space from Fréchet and $ X $ to be a metric space for which there is a continuous surjection $ f: E rightarrow X $. So must $ X $ to be homeomorphic to a Fréchet space or a Riemannian manifold of finite dimension.

I was trying to show that:

$$ n ^ {p} = sum_ {k = 0} ^ {n} binom {n} {k} S_ {p, k} $$

or $ S_ {p, k} $ : is the number of the overjection of a set of p elements to a set of n elements,

I show up to now that:

$$ S_ {p, n} = n (S_ {p-1, n-1} + S_ {p-1, n}) $$

I'm trying to induce but that does not seem to work.

A little help please!

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