Show that if ##f:A\rightarrow B## is surjective then....

[Potatochip911]

Homework Statement

Show that if ##f:A\rightarrow B## is subjective and ##H\subseteq B## then ##f(f^{-1}(H))=H##, give an example to show the equality need to hold if ##f## if not surjective.

Homework Equations

3. The Attempt at a Solution [/B]

I know that I want to show that an element ##x\in H## by starting from the LHS but I'm not really sure what to do. In a similar proof in my notes we had a step where we did something like this:

Let ##x\in f(f^{-1}(H))=H## then ##\exists y\in f^{-1}(H)## such that ##x=f(y)##, I'm not really sure where to go from here or what the point of this step is. I'm also confused by ##f^{-1}(H)##, if I'm understanding this question correctly ##H## is the domain of the function so what exactly is ##f^{-1}(H)## and why is letting a ##y## exist in this useful? And lastly I'm not sure how the definition of a surjective function is useful:
A function ##f## is said to be surjective (or map A onto B) if ##f(A)=B##, that is if the range ##R(f)=B##.

 

[S.G. Janssens]

Homework Statement

Show that if f:A→Bf:A\rightarrow B is subjective and H⊆BH\subseteq B then f(f−1(H))=Hf(f^{-1}(H))=H, give an example to show the equality need to hold if ff if not surjective.

"subjective" should be "surjective" and "need to hold" should be "need not hold".

Let ##x \in f(f^{-1}(H)) = H##

The equality is what you need to prove, so you cannot write it down yet.

I'm also confused by ##f^{-1}(H)##, if I'm understanding this question correctly ##H## is the domain of the function so what exactly is ##f^{-1}(H)##

##A## is the domain of the function, ##B## is the codomain of the function and ##H## is nothing more than a subset of ##B##. The notation ##f^{-1}(H)## simply means, by definition,
$$
f^{-1}(H) := \{x \in A\,:\, f(x) \in H\}
$$
With this in place, you are asked to prove that ##f(f^{-1}(H)) = H##. For this you should prove two inclusions: ##f(f^{-1}(H)) \subseteq H## and ##f(f^{-1}(H)) \supseteq H##. Why don't you start with the first inclusion?

 

[Potatochip911]

"subjective" should be "surjective" and "need to hold" should be "need not hold".
The equality is what you need to prove, so you cannot write it down yet.
##A## is the domain of the function, ##B## is the codomain of the function and ##H## is nothing more than a subset of ##B##. The notation ##f^{-1}(H)## simply means, by definition,
$$
f^{-1}(H) := \{x \in A\,:\, f(x) \in H\}
$$
With this in place, you are asked to prove that ##f(f^{-1}(H)) = H##. For this you should prove two inclusions: ##f(f^{-1}(H)) \subseteq H## and ##f(f^{-1}(H)) \supseteq H##. Why don't you start with the first inclusion?

Yea I'm not quite sure why I wrote subjective there since right after that I wrote surjective. I also accidentally tagged on the ##=H## to ##x\in f(f^{-1}(H))##. Going back to the proof:

Let ##x\in f(f^{-1}(H))## then ##\exists y\in f^{-1}(H)## such that ##x=f(y)##, now I want to show that ##x\in H## using the fact that ##f:A\rightarrow B## is surjective but I can't seem to figure it out. I'm not sure if this is useful or not but if ##\exists y\in f^{-1}(H)## then it seems logical that ##f(y)\in H## but still not sure where to go from here.

 

[vela]

Let ##x\in f(f^{-1}(H))## then ##\exists y\in f^{-1}(H)## such that ##x=f(y)##. Now I want to show that ##x\in H## using the fact that ##f:A\rightarrow B## is surjective but I can't seem to figure it out.

You don't need to use the fact that f is surjective for this part.

 

[Potatochip911]

You don't need to use the fact that f is surjective for this part.

Is ##f(y)\in H## going in the right direction or do I need to think of something else entirely?

 

[Fredrik]

Is ##f(y)\in H## going in the right direction or do I need to think of something else entirely?

Yes, it's the correct way to proceed. If x=f(y), and f(y) is an element of H, then x is...

Note that what you're using to justify that step is just the definition of the notation ##f^{-1}(H)##.

 

[Potatochip911]

Yes, it's the correct way to proceed. If x=f(y), and f(y) is an element of H, then x is...

Note that what you're using to justify that step is just the definition of the notation ##f^{-1}(H)##.

Oh so since ##x=f(y)## and from ##\exists y\in f^{-1}(H)## such that ##f(y)\in H## then ##\exists h\in H## such that ##x=h## which implies ##x\in H##?

 

[Fredrik]

Right, but it's a bit of a detour to bring in a new variable h. Also, (this is a bit of a nitpick) when you have mentioned that "there's a ##y\in f^{-1}(H)## such that...", then it's a bit weird to talk about the properties of y in the next sentence. The problem is that the "there exists" makes y a dummy variable; it can be replaced by any other variable without changing the meaning of the statement. So (very) strictly speaking, it's not clear what y the next sentence is referring to.

This is how I would write the part of the proof that you have completed so far:

Let ##x\in f(f^{-1}(H))##. Let ##y\in f^{-1}(H)## be such that ##f(y)=x##. (Such a ##y## exists because ##x\in f(f^{-1}(H))##). Since ##y\in f^{-1}(H)##, we have ##f(y)\in H##. Since ##x=f(y)## and ##f(y)\in H##, we have ##x\in H##.

 

[Potatochip911]

Right, but it's a bit of a detour to bring in a new variable h. Also, (this is a bit of a nitpick) when you have mentioned that "there's a ##y\in f^{-1}(H)## such that...", then it's a bit weird to talk about the properties of y in the next sentence. The problem is that the "there exists" makes y a dummy variable; it can be replaced by any other variable without changing the meaning of the statement. So (very) strictly speaking, it's not clear what y the next sentence is referring to.

This is how I would write the part of the proof that you have completed so far:

Let ##x\in f(f^{-1}(H))##. Let ##y\in f^{-1}(H)## be such that ##f(y)=x##. (Such a ##y## exists because ##x\in f(f^{-1}(H))##). Since ##y\in f^{-1}(H)##, we have ##f(y)\in H##. Since ##x=f(y)## and ##f(y)\in H##, we have ##x\in H##.

Yea I suppose it is quite odd what I did with ##y## there, I will definitely keep that in mind for future proofs. So now I want to show that ##H\subseteq f(f^{-1}(H))## given that ##f:A\rightarrow B## is subjective and that ##H\subseteq B##:

I'm not too confident about this but it's all I could come up. Let ##x\in H##, ##f:A\rightarrow B## is surjective so we will be using the inverse image definition of ##f^{-1}## i.e. ##f^{-1}(H):=\{ x\in A : f(x)\in H\}##, Since ##f## is surjective there exists an element ##a\in A## for every ##b\in B## so ##a\in f^{-1}(H)## and ##f(a)\in f(f^{-1}(H))##, since ##f(a)\subseteq B## and ##H\subseteq B## we have that ##x\in f(f^{-1}(H))##

 

[Fredrik]

Since ##f## is surjective there exists an element ##a\in A## for every ##b\in B## so ##a\in f^{-1}(H)## and ##f(a)\in f(f^{-1}(H))##, since ##f(a)\subseteq B## and ##H\subseteq B## we have that ##x\in f(f^{-1}(H))##

I can't really follow what you're doing there. The "there exists" makes ##a## a dummy variable in the first part of the sentence, and then you make a claim about ##a##. That claim isn't true for every possible ##a## that corresponds to some ##b\in B##.

I would start with this: You know that ##x\in H\subseteq B## and that ##f:A\to B## is surjective. This should tell you something about ##x## that will be useful.

 

[Potatochip911]

I can't really follow what you're doing there. The "there exists" makes ##a## a dummy variable in the first part of the sentence, and then you make a claim about ##a##. That claim isn't true for every possible ##a## that corresponds to some ##b\in B##.

I would start with this: You know that ##x\in H\subseteq B## and that ##f:A\to B## is surjective. This should tell you something about ##x## that will be useful.

What I'm trying to say (I think it's similar to your suggestion) is that ##x\in H\subseteq B## and since ##f:A\rightarrow B## is surjective then we know that for every ##b\in B## there exists ##a\in A## such that ##f(a)=b## and from the definition of the inverse image we have that ##f^{-1}(H)\in A##. Therefore, there exists and element ##h\in H## such that ##f^{-1}(h)=a\Rightarrow f(f^{-1}(h))=f(a)##. Since ##f(a)=b\in B##, ##H\subseteq B## and using the fact that ##h\in H## we have that ##H\in f(f^{-1}(H)##. Sorry if this is pretty much the exact same thing as I did before.

 

[Fredrik]

What I'm trying to say (I think it's similar to your suggestion) is that ##x\in H\subseteq B## and since ##f:A\rightarrow B## is surjective then we know that for every ##b\in B## there exists ##a\in A## such that ##f(a)=b##

Right. And if something holds "for all b in B", and x is in B, then it holds for x.

and from the definition of the inverse image we have that ##f^{-1}(H)\in A##.

Right.

Therefore, there exists and element ##h\in H## such that ##f^{-1}(h)=a\Rightarrow f(f^{-1}(h))=f(a)##.

Do you mean such that the first equality holds or such that the implication holds? I assume the former. The claim doesn't entirely make sense. You're saying something about ##a##, but ##a## was a dummy variable in an earlier statement and is undefined now. If you're going to make a statement P(a) about a variable ##a##, then you need to do one of the following three things:

1. Assign a value to ##a## first: Define ##a=1##.
2. Make ##a## the target of a "for all": ##\forall a~P(a)##
3. Make ##a## the target of a "there exists": ##\exists a~P(a)##.

If you meant "for all ##a\in A## such that ##f(a)=b##, there's an ##h\in H## such that ##f^{-1}(\{h\})=\{a\}##", then the sentence is true for some choices of ##b## and false for others. (It's false for all b that are not in H).

The notation ##f^{-1}(H)## makes sense even when f is not invertible, but the notation ##f^{-1}(h)## doesn't.

 

[Potatochip911]

Right. And if something holds "for all b in B", and x is in B, then it holds for x.Right.Do you mean such that the first equality holds or such that the implication holds? I assume the former. The claim doesn't entirely make sense. You're saying something about ##a##, but ##a## was a dummy variable in an earlier statement and is undefined now. If you're going to make a statement P(a) about a variable ##a##, then you need to do one of the following three things:

1. Assign a value to ##a## first: Define ##a=1##.
2. Make ##a## the target of a "for all": ##\forall a~P(a)##
3. Make ##a## the target of a "there exists": ##\exists a~P(a)##.

If you meant "for all ##a\in A## such that ##f(a)=b##, there's an ##h\in H## such that ##f^{-1}(\{h\})=\{a\}##", then the sentence is true for some choices of ##b## and false for others. (It's false for all b that are not in H).

The notation ##f^{-1}(H)## makes sense even when f is not invertible, but the notation ##f^{-1}(h)## doesn't.

Yea I meant that the first equality would hold. So it seems like saying there exists an ##h\in H## is a bad idea since h is a single variable? Is that why ##f{^-1}(h)## doesn't make sense?

 

[Fredrik]

##f^{-1}(h)## denotes the output that you get when the function ##f^{-1}## takes ##h## as input. If ##f## isn't invertible, there is no ##f^{-1}## and therefore no ##f^{-1}(h)##. There is however a ##f^{-1}(\{h\})##, because this is a notation for the set ##\{x\in X| f(x)\in\{h\}\}##, which is equal to ##\{x\in X| f(x)=h\}##.

 

[Potatochip911]

##f^{-1}(h)## denotes the output that you get when the function ##f^{-1}## takes ##h## as input. If ##f## isn't invertible, there is no ##f^{-1}## and therefore no ##f^{-1}(h)##. There is however a ##f^{-1}(\{h\})##, because this is a notation for the set ##\{x\in X| f(x)\in\{h\}\}##, which is equal to ##\{x\in X| f(x)=h\}##.

So if I go back to the definition of the inverse part where I say:

From the definition of the inverse we have that ##f^{-1}(H)\in A## so ##\exists a\in A## such that ##\forall a\in f(a)## the following implications hold ##f^{-1}(H)=a \Rightarrow f(f^{-1}(H))=f(a)## and then since ##f:A\to B## is surjective we know that ##A## will map to all elements of ##B## and since ##H\subseteq B## and ##f(a)\subseteq B## we have ##f(a)\subseteq H\Rightarrow f(f^{-1}(H))\in H##

 

[Fredrik]

So if I go back to the definition of the inverse part where I say:

From the definition of the inverse we have that ##f^{-1}(H)\in A##

The definition of "inverse" isn't involved here. ##f^{-1}(H)## is the preimage of ##H## under ##f##, not the image of ##H## under ##f^{-1}##. So the definition you need to use is the definition of "preimage". It should tell you that ##f^{-1}(H)## isn't an element of ##A##.

so ##\exists a\in A## such that ##\forall a\in f(a)## the following implications hold ##f^{-1}(H)=a \Rightarrow f(f^{-1}(H))=f(a)##

It doesn't make sense to have "there exists" and "for all" target the same variable in a single statement. It's impossible to tell which one of these statements is the one you meant to make
\begin{align*}
\exists x\in A\quad \forall y\in f(a)\quad f^{-1}(H)=x \Rightarrow f(f^{-1}(H))=f(x)\\
\exists x\in A\quad \forall y\in f(a)\quad f^{-1}(H)=x \Rightarrow f(f^{-1}(H))=f(y)\\
\exists x\in A\quad \forall y\in f(a)\quad f^{-1}(H)=y \Rightarrow f(f^{-1}(H))=f(x)\\
\exists x\in A\quad \forall y\in f(a)\quad f^{-1}(H)=y \Rightarrow f(f^{-1}(H))=f(y)\\
\end{align*}
Unfortunately none of them makes sense. ##f(a)## denotes an element of ##B##, not a subset of ##A## or ##B##. ##f^{-1}(H)## denotes a subset of ##A##, not an element of ##A##. ##f(f^{-1}(H))## is a subset of ##B## and ##f(a)## is an element of ##B##.

 

[Potatochip911]

The definition of "inverse" isn't involved here. ##f^{-1}(H)## is the preimage of ##H## under ##f##, not the image of ##H## under ##f^{-1}##. So the definition you need to use is the definition of "preimage". It should tell you that ##f^{-1}(H)## isn't an element of ##A##.It doesn't make sense to have "there exists" and "for all" target the same variable in a single statement. It's impossible to tell which one of these statements is the one you meant to make
\begin{align*}
\exists x\in A\quad \forall y\in f(a)\quad f^{-1}(H)=x \Rightarrow f(f^{-1}(H))=f(x)\\
\exists x\in A\quad \forall y\in f(a)\quad f^{-1}(H)=x \Rightarrow f(f^{-1}(H))=f(y)\\
\exists x\in A\quad \forall y\in f(a)\quad f^{-1}(H)=y \Rightarrow f(f^{-1}(H))=f(x)\\
\exists x\in A\quad \forall y\in f(a)\quad f^{-1}(H)=y \Rightarrow f(f^{-1}(H))=f(y)\\
\end{align*}
Unfortunately none of them makes sense. ##f(a)## denotes an element of ##B##, not a subset of ##A## or ##B##. ##f^{-1}(H)## denotes a subset of ##A##, not an element of ##A##. ##f(f^{-1}(H))## is a subset of ##B## and ##f(a)## is an element of ##B##.

Okay so I should just define the preimage as ##f^{-1}(H)=\{x\in A: f(x)\in H\}##; ##f:A\to B## is surjective and we want to show that ##H\subseteq f(f^{-1}(H))##, Let ##x\in H##, then after applying the preimage we get ##f^{-1}(x)\in f^{-1}(H)##, and then multiplying by the function we get ##f(f^{-1}(H))\in f^{-1}(H)##, I reread the entire thread again and there's something I don't understand from the first proof.

Let ##x\in f(f^{-1}(H))##. Let ##y\in f^{-1}(H)## be such that ##f(y)=x##. (Such a ##y## exists because ##x\in f(f^{-1}(H))##). Since ##y\in f^{-1}(H)##, we have ##f(y)\in H##. Since ##x=f(y)## and ##f(y)\in H##, we have ##x\in H##.

Can you explain why such a ##y## exists because ##x\in f(f^{-1}(H))##? Is it because if ##y\in f^{-1}(H)## can we write ##x\in f(f^{-1}(H))\Rightarrow x\in f(y)##?

 

[Fredrik]

Let ##x\in H##, then after applying the preimage we get ##f^{-1}(x)\in f^{-1}(H)##,

This isn't correct. ##f^{-1}(x)## exists only if ##f## is invertible. Since the problem doesn't tell you that ##f## is injective, the notation ##f^{-1}(x)## (where x is an element of B rather than a subset of B), shouldn't appear anywhere in the proof.

and then multiplying by the function we get ##f(f^{-1}(H))\in f^{-1}(H)##

Is there a typo here? Did you mean ##f(f^{-1}(x))\in H##? If ##f## is invertible and ##f^{-1}(x)\in f^{-1}(H)##, then it's correct to conclude that ##f(f^{-1}(x))\in H##, but the correct justification is that it follows immediately from the definition of ##f^{-1}(H)##. You should make sure that you understand why.

I reread the entire thread again and there's something I don't understand from the first proof.

Can you explain why such a ##y## exists because ##x\in f(f^{-1}(H))##? Is it because if ##y\in f^{-1}(H)## can we write ##x\in f(f^{-1}(H))\Rightarrow x\in f(y)##?

##x## is an arbitrary element of ##f(f^{-1}(H))##. By definition of the image of a set under f, we have ##f(f^{-1}(H))=\{f(y)|y\in A\}##. The right-hand side is an abbreviated notation for ##\{z\in B|\exists y\in A~f(y)=z\}##.

If you prefer the abbreviated notation: Since ##x\in\{f(y)|y\in A\}##, there's a ##y\in A## such that ##f(y)=x##.

If you prefer the full notation: Since ##x\in\{z\in B|\exists y\in A~f(y)=z\}##, there's a ##y\in A## such that ##f(y)=x##.

 

[Potatochip911]

Is there a typo here? Did you mean f(f−1(x))∈Hf(f^{-1}(x))\in H? If ff is invertible and f−1(x)∈f−1(H)f^{-1}(x)\in f^{-1}(H), then it's correct to conclude that f(f−1(x))∈Hf(f^{-1}(x))\in H, but the correct justification is that it follows immediately from the definition of f−1(H)f^{-1}(H). You should make sure that you understand why.

Yep I accidentally put the ##f^{-1}## there.

Let ##y\in H##, from the definition of the preimage we have that ##f^{-1}(H):=\{a\in A: f(a)\in H\}## and since the function is surjective we have that ##\forall h\in H \exists a\in A## such that ##f(a)=h##. Therefore, there's an element ##a\in A## such that ##f(a)=y##, so we have ##a\in A## and ##f(a)\in H## and from ##f(a)=y\Rightarrow f(a)\in H##, so we have shown that ##y\in f^{-1}(H)##, Is this correct? Obviously this isn't the entire proof since I have to show ##y\in f(f^{-1}(H))## but it seems to at least be getting me somewhere.

 

[Fredrik]

Let ##y\in H##, from the definition of the preimage we have that ##f^{-1}(H):=\{a\in A: f(a)\in H\}## and since the function is surjective we have that ##\forall h\in H \exists a\in A## such that ##f(a)=h##. Therefore, there's an element ##a\in A## such that ##f(a)=y##,

This part is excellent.

so we have ##a\in A## and ##f(a)\in H## and from ##f(a)=y\Rightarrow f(a)\in H##, so we have shown that ##y\in f^{-1}(H)##, Is this correct?

It is not. It looks like you used the result you found (there's an ##a\in A## such that ##f(a)=y##) only to "prove" the assumption that you started with (##y\in H##), and then jumped to the conclusion that ##y\in f^{-1}(H)##. Note that H is a subset of B, and ##f^{-1}(H)## is a subset of ##A##.

Let ##y\in H##. Let ##a\in A## be such that ##f(a)=y##. (You have explained why such an ##a## must exist). To make progress from this point, you need to find a better way to use that ##f(a)=y\in H##.

 

[Potatochip911]

This part is excellent.It is not. It looks like you used the result you found (there's an ##a\in A## such that ##f(a)=y##) only to "prove" the assumption that you started with (##y\in H##), and then jumped to the conclusion that ##y\in f^{-1}(H)##. Note that H is a subset of B, and ##f^{-1}(H)## is a subset of ##A##.

Let ##y\in H##. Let ##a\in A## be such that ##f(a)=y##. (You have explained why such an ##a## must exist). To make progress from this point, you need to find a better way to use that ##f(a)=y\in H##.

So I am now thinking something like this. We have ##a\in A## and ##f(a)\in H##, since ##f(a)=y\in H##. There exists an element ##h\in H## such that ##f^{-1}(\{h\})=a##. Therefore from ##f(a)=y## we have ##f(a)=f(f^{-1}(\{h\}))=y\in H##

 

[Fredrik]

So I am now thinking something like this. We have ##a\in A## and ##f(a)\in H##, since ##f(a)=y\in H##. There exists an element ##h\in H## such that ##f^{-1}(\{h\})=a##.

It's possible that there exists an ##h\in H## such that ##f^{-1}(\{h\})=\{a\}## (not ##=a##), but you can't know that for sure, since we haven't assumed that ##f## is injective.

Keep in mind that the statement that you're trying to prove is ##H\subseteq f(f^{-1}(H))##. This suggests that you should be looking for true statements that involve ##f^{-1}(H)##.

 

[Potatochip911]

It's possible that there exists an ##h\in H## such that ##f^{-1}(\{h\})=\{a\}## (not ##=a##), but you can't know that for sure, since we haven't assumed that ##f## is injective.

Keep in mind that the statement that you're trying to prove is ##H\subseteq f(f^{-1}(H))##. This suggests that you should be looking for true statements that involve ##f^{-1}(H)##.

Is it possible to assume that there's "##a\in A## such that ##\{a\}=f^{-1}(\{H\})##?" I feel like that's not possible since if the function is surjective then every single element in B could have more than one element in A mapping to it. I'm guessing I have to use the fact then that ##f^{-1}(\{H\}):=\{x\in A:f(x)\in H\}## so by the definition of the inverse image ##f^{-1}(\{H\})\in A##. Now if there exists an element ##x\in f^{-1}(\{H\})## then from the definition of ##f^{-1}(\{H\}\Rightarrow f(x)\in f(f^{-1}(\{H\})## and we also have that ##f(x)\in H## from the definition. So we get the result that ##f(x)\in H\in f(f^{-1}(\{H\})##.

 

[Fredrik]

Is it possible to assume that there's "##a\in A## such that ##\{a\}=f^{-1}(\{H\})##?" I feel like that's not possible since if the function is surjective then every single element in B could have more than one element in A mapping to it.

H is a subset of B, but ##\{H\}## isn't, so the notation ##f^{-1}(\{H\})## is undefined.

If you meant to write ##f^{-1}(H)##, there's no reason to think that such an ##a## exists. Such an ##a## would exist e.g. if we had assumed that f is injective and that H is a singleton set (i.e. a set with only one element).

If you meant to write ##f^{-1}(h)##, where h is some element of H, then there's still no reason to think that such an ##a## exists. Such an ##a## would exist e.g. if we had assumed that f is injective.

I'm guessing I have to use the fact then that ##f^{-1}(\{H\}):=\{x\in A:f(x)\in H\}##

You will have to use that ##f^{-1}(H)## is equal to that right-hand side.

You will also have to use the definition of ##f(E)## for an arbitrary ##E\subseteq A##.

so by the definition of the inverse image ##f^{-1}(\{H\})\in A##.

The term is "preimage", and (regardless of whether you meant H or h) it's ##\subseteq##, not ##\in##.

Now if there exists an element ##x\in f^{-1}(\{H\})## then from the definition of ##f^{-1}(\{H\}\Rightarrow f(x)\in f(f^{-1}(\{H\})## and we also have that ##f(x)\in H## from the definition. So we get the result that ##f(x)\in H\in f(f^{-1}(\{H\})##.

If you meant that ##x\in f^{-1}(H)## implies that ##f(x)\in f(f^{-1}(H))##, then you're right.

 

[Potatochip911]

You will have to use that ##f^{-1}(H)## is equal to that right-hand side.

You will also have to use the definition of ##f(E)## for an arbitrary ##E\subseteq A##.The term is "preimage", and (regardless of whether you meant H or h) it's ##\subseteq##, not ##\in##.If you meant that ##x\in f^{-1}(H)## implies that ##f(x)\in f(f^{-1}(H))##, then you're right.

I'm quite lost as to what to do with the sets but I think I've found a way to prove it. Let ##y\in H##, since ##f:A\to B## is surjective ##\exists a\in A## such that ##f(a)=y## therefore, ##a\in f^{-1}(H)## where ##f^{-1}(H):=\{a\in A:f(a)\in H\}##. From this we get that ##f(a)\in f(f^{-1}(H))## and we know that ##f(a)=y## and that ##y\in H## so ##H\subseteq f(f^{-1}(H))##