Let $f: \mathbb{R} → \mathbb{R}$ be a function and $c$ a nonzero real number. If $f$ is onto, is $c · f$ also onto? Use the following definition: If $f : \mathbb{R} → \mathbb{R}$ is a function and c is a nonzero real number, the function $(c·f): \mathbb{R} → \mathbb{R}$ is defined by the formula $(c· f )(x) = c· f (x)$ for all real numbers $x$.
To prove
Let $f:\mathbb{R} → \mathbb{R}$ be a function and $c$ a nonzero real number. If $f$ is onto, is $c · f$ also onto? 
Here is my solution. Is it correct? My professor gave an answer which is shorter. But I want to know whether my answer is correct.
 A: Here's a tip that will help you in general with proofs.
Even if you don't know why this would be true, the form of the argument is automatic.  First, write the hypothesis - the ground statement that you are assuming to be true before the argument can start.  In this case it's

Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ is surjective. Suppose that $c \in \mathbb{R}$ is nonzero and define $g: \mathbb{R} \rightarrow \mathbb{R}$ by $g(x) = c \cdot f(x)$.

It's not completely necessary to define the symbol $g$ instead of just using $(c \cdot f)$ but personally I think it will make the argument clearer to yourself and your reader by giving some separation between what you know about $f$ and what you know about $g$.
You're trying to prove that $g(x)$ is surjective. By definition that means
$$\forall y \in \mathbb{R} \, \exists x \in \mathbb{R}\, g(x) = y.$$
The form of this symbolic sentence tells you exactly how to start the proof.  It's a fill-in-the-blank argument.  The $\forall y \in \mathbb{R}$ means that there is some completely arbitrary real number $y$ that you have to prove something about.  So start by writing

Let $y \in \mathbb{R}$.

Next, you see $\exists x \in \mathbb{R}$.  This means that you have to choose a particular x to do something you like.  We don't know how to choose it yet, so let's write

Choose $x$ _____________.

We'll fill in the blank later, when we figure out what $x$ needs to be.  Note that the order is important - we have to introduce the symbol $y$ before we introduce the symbol $x$, because that's the order of the variables in the sentence we're trying to prove.
Now we have to get to the conclusion somehow.  It's okay if you don't know how to get there yet.  We know our ending anyway: we will have $g(x) = y$.

Then ________ and so $g(x) = y$.

We can fill in the blanks here a bit, because we know what $g(x)$ is.  It will be something like

Then $g(x) = c \cdot f(x) = $____ $= y$.  So $g(x) = y$.

Putting it all together, we have the bare bones of our proof:

Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ is surjective. Suppose that $c \in \mathbb{R}$ is nonzero and define $g: \mathbb{R} \rightarrow \mathbb{R}$ by $g(x) = c \cdot f(x)$. 
Let $y \in \mathbb{R}$. Choose $x$ _____________.  Then $g(x) = c \cdot f(x) = $____ $= y$.  So $g(x) = y$.  Thus $g$ is surjective.

Now you have the logical structure mostly in place, so you just have to think of the actual argument.  How can you choose $x$?  Note that you're allowed to define $x$ in terms of $y$, since $y$ was introduced before $x$.  Also, keep track of our premises: We are assuming $f$ is surjective, but we have not used that fact yet.  So, you need to find a way to use our knowledge that $f$ is surjective to prove that $g$ is surjective.
I'll let you think about how to finish.
A: When you introduced $t$ in your solution, you defined it in terms of some $y$ you had already fixed. This is your mistake. You need to show that given arbitrary $t\in\Bbb R$, there exists some $x$ with $(c\cdot f)(x)=t$. Here's the way you should approach it:
Let $y\in\Bbb R$ be arbitrary. Then $\frac yc\in\Bbb R$, and by assumption there exists some $x\in\Bbb R$ with $f(x)=\frac yc$. Multiplying both sides by $c$, we see that $y=(c\cdot f)(x)$. Since $y$ was arbitrary, we conclude that $c\cdot f$ is onto.
A: Let $y\in \mathbb R$ and $c\neq 0$.
$\;\;\;\;\;\frac {y}{c}\in\mathbb R $ and $f $ is onto 
$$\implies $$
$$\exists x\in\mathbb R\;\;:\; f (x)=\frac {y}{c}$$
$$\implies $$
$$\exists x\in\mathbb R\;\;:\;cf (x)=(c.f)(x)=y $$
$$\implies $$
$\;\;\;\;cf $ is onto.
A: The sentence "Then $t$ is a real number since $y$ is a real number and $c$ is a nonzero real number" in your solution is not correct.  To show that $cf: \mathbb{R} \rightarrow \mathbb{R}$ is onto, pick any point $t$ in the codomain.  We need to show that there is some $x$ in the domain such that $cf$ maps $x$ to $t$. (Note that we do not conclude (like you did) that $t$ is in the codomain.  We assume $t$ is a point in the codomain.)   Also, you need to use the fact that because $c$ is nonzero, we can divide by $c$.  
Observe that since $c \ne 0$, $t/c$ is well-defined and is in the codomain of $f$.  Hence, there exists an $x$ in the domain such that $f$ maps $x$ to $t/c$.  Clearly $cf$ maps this $x$ to $t$.
