LaTeX My sketchy proof on functions and mappings. I( Latex Fixed)

[╔(σ_σ)╝]

Homework Statement

f : X \rightarrow Y , g : Y \rightarrow Z and B \subset Z
Prove that
\left(g \circ f\right)^{-1}\left(B \right) = f^{-1} \left({g^{-1} \left(B\right)\right).What is wrong with this proof ?

The Attempt at a Solution

x_{0} \in \left(g \circ f\right)^{-1}\left(B\right) \Rightarrow g \left(f\left(x_{0} \right)\right)\in B \Rightarrow f \left(x_{0}\right) \in g^{-1} \left(B\right)
f \left(x_{0} \right) \in g^{-1} \left(B\right) \Rightarrow x_{0} \in f^{-1} \left(g^{-1} \left(B \right) \right)

Thus,
\left(g \circ f\right)^{-1}\left(B \right) \subset f^{-1} \left({g^{-1} \left(B \right)\right).

I feel uneasy about the proof. I believe my inferences are correct but there is something unsettling about what I did.

Part 2.

Suppose x_{0} \in f^{-1}\left(g^{-1}\left(B\right)\right)

x_{0}\in f^{-1}\left(g^{-1} \left(B\right)\right) \Rightarrow f \left(x_{0}\right) \in g^{-1}\left(B\right) \Rightarrow g\left(f\left(x_{0}\right)\right) \in B

g \left(f \left(x_{0} \right) \right) \in B \Rightarrow x_{0} \in \left(g \circ f\right)^{-1} \left(B \right)
Thus,

f^{-1} \left({g^{-1} \left(B\right)\right) \subset \left(g \circ f\right)^{-1}\left(B\right).

Therefore,

\left(g\circ f\right)^{-1}\left(B\right) = f^{-1}\left({g^{-1}\left(B\right)\right).Some feedback would be appreciated. I would like to know if this proof is valid and good enough.

 

[Office_Shredder]

You need to put a space between latex commands and the next character.

For example \circg will not show up. It needs to be \circ g

Based on your title maybe you made this change? At any rate it's not coming through on my computer

 

[Mark44]

Fixed some more of your LaTeX. The codomains of the functions weren't showing, plus a few other problems.

Homework Statement

f : X\rightarrow Y , g : Y\rightarrow Z and B\subset Z
Prove that
\left(g\circ f\right)^{-1}\left(B\right) = f^{-1}\left({g^{-1}\left(B\right)\right).


What is wrong with this proof ?

The Attempt at a Solution

x_{0}\in \left(g\circ f\right)^{-1}\left(B\right) \Rightarrow g\left(f\left(x_{0}\right)\right)\in B \Rightarrow f\left(x_{0}\right)\in g^{-1}\left(B\right)
f\left(x_{0}\right)\in g^{-1}\left(B\right) \Rightarrow x_{0}\in f^{-1}\left(g^{-1}\left(B\right)\right)

Thus,
\left(g\circ f\right)^{-1}\left(B\right) \subset f^{-1}\left({g^{-1}\left(B\right)\right).

I feel uneasy about the proof. I believe my inferences are correct but there is something unsettling about what I did.

Part 2.

Suppose x_{0}\in f^{-1}\left(g^{-1}\left(B\right)\right)

x_{0}\in f^{-1}\left(g^{-1}\left(B\right)\right) \Rightarrow f\left(x_{0}\right) \in g^{-1}\left(B\right) \Rightarrow g\left(f\left(x_{0}\right)\right) \in B

g\left(f\left(x_{0}\right)\right) \in B \Rightarrow x_{0} \in \left(g \circ f\right)^{-1}\left(B\right)
Thus,

f^{-1}\left({g^{-1}\left(B\right)\right) \subset \left(g\circ f\right)^{-1}\left(B\right).

Therefore,

\left(g\circ f\right)^{-1}\left(B\right) = f^{-1}\left({g^{-1}\left(B\right)\right).


Some feedback would be appreciated. I would like to know if this proof is valid and good enough.

 

[╔(σ_σ)╝]

The funny thing is that I fixed it all and checked to see if it was okay. When I logged back to read the replies it was unchanged.

The forum always acts weird when I use my phone to post threads. I think the mobile version of the forum has some bugs.

Thanks Mark.Any feedback ?

 

[╔(σ_σ)╝]

Does anyone have any suggestions are comments ?

I would like to know if this proof supplied here is valid . I would like to know if I have to make adjustments or rewrite the entire thing.

 

[logarithmic]

Does anyone have any suggestions are comments ?

I would like to know if this proof supplied here is valid . I would like to know if I have to make adjustments or rewrite the entire thing.

Yes, your proof is correct. It's not dodgy.

I think that the reason you think it may possibly be dodgy (correct me if I'm wrong) is because you keep using the fact that x\in h^{-1}(A) is equivalent to h(x)\in A.

The notation h^{-1} when applied to a set (instead of an element) is not the inverse function, it's the preimage, and so that statement is actually true by definition

 

[╔(σ_σ)╝]

Great. You are correct about my uneasiness.

I thought I was making some claims that may put me into trouble.