Showing that a composite function is bijective. Can i straight away claim that $\varphi : T \rightarrow X$ is injective, and if $f:X \rightarrow Y$ is injective, then this means $f \circ \varphi : T \rightarrow Y$ is injective for as well any set T base on the fact that the composition of two injective functions is injective?
Suppose $f \circ \varphi : a = f \circ \varphi : b$ 
Let $\varphi : a = c$ and $\varphi : b = d$
so $f(c) = f(d)$
Since $f:X \rightarrow  Y$ is injective implies $f(c) = f(d)$ We know that $c = d$.
This means that $f(c) = f(d)$
Since $\varphi : T \rightarrow  X$ is injective implies  $\varphi : a = \varphi : b$. We know that $a = b$.
Therefore i have shown that $f \circ \varphi : a = f \circ \varphi : b$  then $a = b$.
Is my prove implies both direction? Meaning the composite function is injective if and only if $f:X \rightarrow Y$ is injective and $\varphi : T \rightarrow X$ is injective.
How do i show that $\psi \circ f : X \rightarrow T$ is injective if and only if $f:X \rightarrow Y$ is surjective for any set $T$?
 A: Lemma: The composition of two injective functions is injective.
Proof: Let $ X, Y, Z $ be sets and $ f : X \rightarrow Y, g: Y \rightarrow Z $ be injective maps between them. Then
$$ a = b \iff f(a) = f(b) ~\text{for all}~ a, b \in X $$
and
$$ c = d \iff g(c) = g(d) ~\text{for all}~ c,d \in Y $$
in particular, as $ f(a), f(b) \in Y $,
$$ f(a) = f(b) \iff g(f(a)) = g(f(b))$$
so that
$$ a=b \iff f(a) = f(b) \iff g(f(a)) = g(f(b)), ~\text{for all}~ a,b \in X $$
which gives the definition of injectivity for $ g \circ f : X \rightarrow Z $.
As to the last part, what is $ \psi $?
A: The composition of two injective functions is certainly injective (Will's answer has a proof), and the composition of two surjective functions is also surjective. So, if the first function is injective, and the second is bijective when restricted to the image of the first function, their composition will be a bijection.
A: To the first question: No, $f$ does not have to be injective,what you have is 
If $f \circ \phi$ is injective then $\phi$ is injective. 
To show this notice that  $\phi(a)= \phi(b) \Rightarrow f(\phi(a)) = f(\phi(b))  \Rightarrow f\circ \phi(a)= f\circ \phi(b) \Rightarrow a=b $, because $f \circ \phi$ is injective. That means that $\phi$ is injective. 
An example where $f$ is not injective is: 
Take $X=\lbrace 1,2,3,4 \rbrace,Y=\lbrace 1,2,3,4 \rbrace\  \text{and}\ \ T=\lbrace 1,2 \rbrace$.Take $\phi$ to be the identity function and define $f(1) = 1, f(2) =2,f(3)=f(4)=3$. Then you have that $f \circ \phi$ is injective but $f$ is not. 
To the second question: 
As we have shown above, if $\psi \circ f$ is injective you have that $f$ is injective, if $f$ is also surjective then you garantee that $\psi$ is injective. 
