Show that if U and T are onto, then UT is also onto

[Mr Davis 97]

Homework Statement

Same as title, where T maps from a set V to W, and U maps from W to Z.

Homework Equations

The Attempt at a Solution

I know that the standard definition for a function being onto is that for all elements w in the codomain, there exists an element v in the domain such that T(v) = w. I am not sure exactly how to concretely apply this to show that U composed with T is also onto.

To start, we have U(T(v)) = z, and we need to show that for all z in Z there exists a v in V such that that mapping is true. First we can use the fact that U is onto, which means there is always a w = T(v) that maps to z, and since T is onto, there is always a v that maps to w. Is this sufficient to show that UT is also onto? Is there a way to write this argument using less words?

 

[fresh_42]

To start, we have U(T(v)) = z, and we need to show that for all z in Z there exists a v in V such that that mapping is true. First we can use the fact that U is onto, which means there is always a w = T(v) that maps to z, and since T is onto, there is always a v that maps to w. Is this sufficient to show that UT is also onto? Is there a way to write this argument using less words?

You confused the order a bit.

To start let's see what we have (which is always a good idea): $\;V \stackrel{T}{\twoheadrightarrow} W \stackrel{U}{\twoheadrightarrow} Z$.
A map $f: X \twoheadrightarrow Y$ is surjective, if the following is true (you have already said this, but you asked for a short way to write it):
$$ \forall \; y \in Y \, : \, f^{-1}(y) = \{x \in X\,\vert \,f(x)=y\} \neq \emptyset \stackrel{or}{\Longleftrightarrow} \forall \; y \in Y \; \exists \; x\in X \, : \, f(x)=y$$
Now we have to start with an arbitrary point $z \in Z$. The $v\in V$ for which $U(T(v))=z$ holds, has to be shown to exist.

So first, you have to get back from $Z$ to $W$ by $U\; -\;$ no $v$, no $V$ and no $T$ at this stage, because all you have is an arbitrary element $z \in Z$ for which you need to show $(U \circ T)^{-1}(z) \neq \emptyset \; .$
If you have your element in $W$, then you can proceed using $T$, but not earlier.

 

[Mr Davis 97]

But U is surjective, so there must be a w in W s.t. U(w) = z, for any z in Z

 

[fresh_42]

But U is surjective, so there must be a w in W s.t. U(w) = z, for any z in Z

Yes, and for those $w$ you can repeat the argument now with $T$.

 

[Mr Davis 97]

So, basically, we need to show that for all z in Z there exists a v in V s.t. UT(v) = U(T(v)) = z, and we do this by first noting that there must be a w in W s.t. U(w) = z, and also that there must be a v in V s.t. T(v) = w, and hence there must be a v in V s.t. UT(v) = z? For my purposes is this an okay way of putting the argument?

 

[fresh_42]

So, basically, we need to show that for all z in Z there exists a v in V s.t. UT(v) = U(T(v)) = z, and we do this by first noting that there must be a w in W s.t. U(w) = z, and also that there must be a v in V s.t. T(v) = w, and hence there must be a v in V s.t. UT(v) = z? For my purposes is this an okay way of putting the argument?

Yes. I would drop the word surjective in both cases to justify the existence of those elements, but yes, that's it.