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Korisnik
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I'm learning maths myself, but I'm going to university in 2 months. This is my first try at proving anything.
Prove that the composition of surjective functions is also a surjection.
A definition of surjective function:
If [itex]f:S_1\rightarrow S_2[/itex] is a surjection:
[itex]\forall y\in S_2\exists x\in S_1:y=f(x)[/itex].
A definition of composition of functions:
If [itex]f:S_1\rightarrow S_2[/itex] and [itex]g:S_2\rightarrow S_3[/itex] are functions, then [itex]h:S_1\rightarrow S_3[/itex] is the composition of functions f and g:
[itex]g\circ f(x)=g[f(x)]=g(y)=z[/itex].
I've read some of the texts that talk about proofs, and all of them made sense to me, for example the proving of an irrational number and square roots, but I can't say if this proof is "enough".
If [itex]f:S_1\rightarrow S_2[/itex] and [itex]g:S_2\rightarrow S_3[/itex] are surjections by definitions:
[itex]\forall y\in S_2\exists x\in S_1:y=f(x)\ \ \ \ (1)[/itex]
[itex]\forall z\in S_3\exists y\in S_2:z=g(y)\ \ \ \ (2)[/itex],
composition of these two functions [itex]g\circ f(x)[/itex] is also a surjection.
We have to prove:
[itex]\forall z\in S_3\exists x\in S_1:z=g\circ f(x)[/itex].
Let [itex]z\in S_3[/itex], then there is a particular [itex]y\in S_2[/itex] such that [itex]z=g(y)[/itex] by the [itex](2)[/itex] axiom and there is a particular [itex]x\in S_1[/itex] such that [itex]y=f(x)[/itex] by the [itex](1)[/itex] axiom. It follows that [itex]z=g(y)=g[f(x)]=g\circ f(x)[/itex] which is by the definition, the composite of functions [itex]f[/itex] and [itex]g[/itex]. This means that for all [itex]z[/itex] we've found a particular [itex]x[/itex] for which the statement is true.
Homework Statement
Prove that the composition of surjective functions is also a surjection.
Homework Equations
A definition of surjective function:
If [itex]f:S_1\rightarrow S_2[/itex] is a surjection:
[itex]\forall y\in S_2\exists x\in S_1:y=f(x)[/itex].
A definition of composition of functions:
If [itex]f:S_1\rightarrow S_2[/itex] and [itex]g:S_2\rightarrow S_3[/itex] are functions, then [itex]h:S_1\rightarrow S_3[/itex] is the composition of functions f and g:
[itex]g\circ f(x)=g[f(x)]=g(y)=z[/itex].
The Attempt at a Solution
I've read some of the texts that talk about proofs, and all of them made sense to me, for example the proving of an irrational number and square roots, but I can't say if this proof is "enough".
If [itex]f:S_1\rightarrow S_2[/itex] and [itex]g:S_2\rightarrow S_3[/itex] are surjections by definitions:
[itex]\forall y\in S_2\exists x\in S_1:y=f(x)\ \ \ \ (1)[/itex]
[itex]\forall z\in S_3\exists y\in S_2:z=g(y)\ \ \ \ (2)[/itex],
composition of these two functions [itex]g\circ f(x)[/itex] is also a surjection.
We have to prove:
[itex]\forall z\in S_3\exists x\in S_1:z=g\circ f(x)[/itex].
Let [itex]z\in S_3[/itex], then there is a particular [itex]y\in S_2[/itex] such that [itex]z=g(y)[/itex] by the [itex](2)[/itex] axiom and there is a particular [itex]x\in S_1[/itex] such that [itex]y=f(x)[/itex] by the [itex](1)[/itex] axiom. It follows that [itex]z=g(y)=g[f(x)]=g\circ f(x)[/itex] which is by the definition, the composite of functions [itex]f[/itex] and [itex]g[/itex]. This means that for all [itex]z[/itex] we've found a particular [itex]x[/itex] for which the statement is true.
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