What is meant by the composition of maps? Question
Hi all, I would like to seek clarification with regards to what does it mean by 'map of post-composition with f.' (Please refer to the image in the link above.)
To my understanding, it means that we have a collection of all maps from T to X and another collection from T to Y and now we are trying to map from every element in Maps(T,X) to Maps (T,Y) so it like mapping a map to map instead.
Would greatly appreciate if anyone could help me in solving this problem. 
Hi I am the original poster of this question. So far I have managed to show the following.
Can anyone help to verify if my working is correct. As for showing <=, my prof suggested letting T be a singleton, can anyone enlighten me the rational on why should we let T be a singleton? 
 A: Let $T$ be a set and let $\phi_1,\phi_2\in\text{Maps}(T,X)$ with $\phi_1\neq\phi_2$.
That means that $\phi_1(t)\neq\phi_2(t)$ for some $t\in T$. 
Then injectivity of $f$ implies that $f(\phi_1(t))\neq f(\phi_2(t))$ hence $f\circ\phi_1\neq f\circ\phi_2$. 
Proved is now:$$f\text{ is injective}\implies\Phi_T\text{ is injective}$$
and this can be done for any set $T$.
Conversely if $f$ is not injective then choose for $T=X$. 
There are elements $x_1,x_2\in X$ with $x_1\neq x_2$ and $f(x_1)=f(x_2)$. 
Now for $\phi$ take the identity on $X$ and let $\psi$ be described by $x\mapsto x$ if $x\neq x_1$ and $x\mapsto x_2$ otherwise. 
Then $\phi\neq\psi$ but $f\circ\phi=f\circ\psi$.
This proves that $\Phi_X$ is not injective.
You can also choose for other sets $T$ to prove this. It is enough if $T$ contains $2$ distinct elements. In one case both are sent to $x_1$, in the other one is sent to $x_1$ and the other to $x_2$.
A: To the question in topic: the composition of maps is another map, defined this way:
$$
(f\circ \phi)(x) \stackrel{\text{def}}= f(\phi(x))
$$
We take first the value of $\phi$ in $x$ and then the value of $f$ in the value of $\phi$, $\phi(x)$. Of course $f$ and $\phi$ must be defined on "compatible" sets, like in your example: codomain of $\phi$ is the domain of $f$ (some would say that it is enough that $\operatorname{codom} \phi \subset \operatorname{dom} f$, which is true in a way, but not from a strict set-theoretic point of view.)
Some hints for the problem: if $f$ is not an injection, then there are $x_1,\ x_2\in X$ such that $f(x_1) = f(x_2)$. Let's say that $T=\{0\}$, so the maps $\phi$ are defined only on one element. Now, how can we define $\phi_1,\ \phi_2\!:\{0\}\to X$ using $x_1$ and $x_2$, so that $f\circ \phi_1 = f\circ\phi_2$, that is: $f(\phi_1(0)) = f(\phi_2(0))$? How about for other sets $T$?
