Preimage proof?

  • Thread starter SMA_01
  • Start date
  • #1
218
0

Homework Statement



Define f:S→T, where B[itex]\subseteq[/itex]T. Let f-1(B)={x[itex]\in[/itex]S:f(x)[itex]\in[/itex]B} be the preimage of B.

Demonstrate that for any such map f, f(f-1(B))=B.

My main question is, would I prove this using set inclusion both ways?

I was going to begin by letting an element be in the preimage of B, and explain what that means, then mapping that element to B. Would this be correct?

I just need a push in the right direction.

Thank you.
 

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,347
1,022

Homework Statement



Define f:S→T, where B[itex]\subseteq[/itex]T. Let f-1(B)={x[itex]\in[/itex]S:f(x)[itex]\in[/itex]B} be the preimage of B.

Demonstrate that for any such map f, f(f-1(B))=B.

My main question is, would I prove this using set inclusion both ways?

I was going to begin by letting an element be in the preimage of B, and explain what that means, then mapping that element to B. Would this be correct?

I just need a push in the right direction.

Thank you.
Generally, yes. To show set equality you need to show inclusion in both directions.
 
  • #3
22,089
3,292
Define f:S→T, where B[itex]\subseteq[/itex]T. Let f-1(B)={x[itex]\in[/itex]S:f(x)[itex]\in[/itex]B} be the preimage of B.

Demonstrate that for any such map f, f(f-1(B))=B.

Just to let you know, but that equality is false in general. You need extra assumptions on f.
 
  • #4
218
0
SammyS- Thank you

Micromass- Yes, the function needs to be surjective, right? I'm not sure why there were no assumptions...
 
  • #5
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,347
1,022
Just to let you know, but that equality is false in general. You need extra assumptions on f.
I'm very reluctant to question you, micromass.

But isn't the extra assumption of f needed for [itex]f^{-1}\left(f(A)\right)=A[/itex] rather than for [itex]f\left(f^{-1}(B)\right)=B\ ?[/itex]
 
  • #6
22,089
3,292
I'm very reluctant to question you, micromass.

But isn't the extra assumption of f needed for [itex]f^{-1}\left(f(A)\right)=A[/itex] rather than for [itex]f\left(f^{-1}(B)\right)=B\ ?[/itex]
An extra assumption is needed for both. Consider [itex]f(x)=x^2[/itex] and B=[-1,0]. Then [itex]f^{-1}(B)=\{0\}[/itex] and thus [itex]f(f^{-1}(B))=\{0\}[/itex]. The problem is that f is not surjective.
 
  • #7
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,347
1,022
An extra assumption is needed for both. Consider [itex]f(x)=x^2[/itex] and B=[-1,0]. Then [itex]f^{-1}(B)=\{0\}[/itex] and thus [itex]f(f^{-1}(B))=\{0\}[/itex]. The problem is that f is not surjective.
Well, my reluctance was well founded!

Thanks as always.
 

Related Threads on Preimage proof?

  • Last Post
Replies
16
Views
3K
  • Last Post
Replies
2
Views
778
  • Last Post
Replies
2
Views
835
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
3K
Replies
4
Views
774
  • Last Post
Replies
1
Views
523
Replies
1
Views
695
Top