Before we start talking about category theory. We will talk about Sets in this post; however, in the next few posts we will generalize the notion to arbitrarily category and give some examples.
Normally, when we talk about a function $f : A \rightarrow B$, the properties of f, whether it is injective or surjective is given by elements. In this section, we will develop a more mature way of thinking about injectivity and surjective, and show that any function has some canonical decomposition. This different mode of thinking will enable us to gain some intuition when we start working with categories.
To introduce this mature way of thinking we need the following theorem:
Theorem 1:
(1)A function $f : A \rightarrow B$ is injective iff it has a left inverse
(2)A function $f : A \rightarrow B$ is surjective iff it has a right inverse.
We can see that theorem 1 allows us to define injectivity, and surjectivity not in terms of elements but in terms of whether a specific map exist. We can do better than this we can define injectivity, and surjectivity by how f interact with other maps, that is one can think of it of how the map f handle itself against other maps.
Definition 1:
A function $f : A \rightarrow B$ is said to be a monomorphic iff for every set $Z$ and for every map $\alpha_1,\alpha_2 : Z \rightarrow B$ such that $f \circ \alpha_1 = f \circ \alpha_2 \implies \alpha_1 = \alpha_2$.
Definition 2:
A function $f : A \rightarrow B$ is said to be a epimorphic iff for every set $U$ and for every map $\beta_1,\beta_2 : B \rightarrow U$ such that $\beta_1 \circ f = \beta \circ f \implies \beta_1 = \beta_2$.
We can see from the above definitions that monomorphic and epimorphic functions are ones that interact with special maps in some way. In the category of sets we can say more than that.
Theorem 2:
1)A function $f : A \rightarrow B$ is injective iff f is monomorphic.
2)A function $f : A \rightarrow B$ is surjective iff f is epimorphic.
Proof:
(1)
Suppose that $f$ is injective. Suppose that Z is a set and $\alpha_1,\alpha_2 : Z \rightarrow A$ such that $f \circ \alpha_1 = f \circ \alpha_2$ since f has a left inverse this mean that $\alpha_1 = \alpha_2$. Thus, f is a monomorphic map.
Conversely, suppose that f is a monomorphic map. Suppose that $a_1,a_2$ are arbitrarily elements of A such that $f(a_1) = f(a_2)$.
Let $Z = \{a_1,a_2\}$ and let $\alpha_1 : Z \rightarrow A$, $\alpha_2 : Z \rightarrow A$ be defined as:
$\alpha_1(a_1) = a_2$ $\alpha_2(a_1) = a_1$
$\alpha_1(a_2) = a_1$ $\alpha_2(a_2) = a_2$.
Then, one can check that $$f \circ \alpha_1 = f \circ \alpha_2$$
Thus $\alpha_1 = \alpha_2$. This means that $a_1 = a_2$, so f is injective.
(2)
Suppose that $f$ is surjective. We can use similar argument as (1) to prove that f is an epimorphic function. Here we use the fact that f has a right inverse.
Conversely, suppose that f is an epimorphic function. We shall argue by contradiction. Assume that f is not surjective, so there exists $b \in B$ such that there is no $a \in A$ that gets mapped to it. Then Suppose we set $U = B$. Suppose we have a fixed map $\beta_1 : B \rightarrow A$. Construct $\beta_2 : B \rightarrow A$ to be the same as $\beta_1$ except at the point $b$. That is $\beta_1(b) \neq \beta_2(b)$. Then we have: $$\beta_1 \circ f = \beta_2 \circ f$$. However, by construction: $$\beta_1 \neq \beta_2$$. This gives us a contradiction. $\square$
Theorem 1 above says that we can think of injectivity and surjectivity by how it interacts with maps. We don't care about elements, or whether it has a specific map which satisfies certain properties. This will be very useful for us later when we explore categories.
The next theorem will tell us that any function has a very specific decomposition. This will be useful to us when we start talking about talking about first isomorphism theorem.
Theorem 2:
Suppose we have an arbitrarily function $f : A \rightarrow B$. Then, f can be decomposed as follows
$$A \twoheadrightarrow A/\sim \rightarrow Im(f) \hookrightarrow B.$$
The equivalence relation $\sim$ is defined as $a_1 \sim a_2 \iff f(a_1) = f(a_2)$. Essentially, what this theorem says is that any function can be realized as a surjection followed by an isomorphism followed by an injection.
Proof:
The first map is the projection map given by $ a \mapsto [a]$. The second map is defined as $[a] \mapsto f(a)$. The third map is just the inclusion map. So, all together we have $a \mapsto [a] \mapsto f(a) \mapsto f(a)$. Therefore, we just need to check if the middle map is an isomorphism. Denote that map by $\theta$.
Suppose that $b \in Im(f)$. Then, by definition there exists $a \in A$ such that $f(a) = b$. Consider the element given by $[a]$. Then, by definition, $$\theta([a]) = f(a) = b.$$ Thus it is surjective.
Suppose that $\theta([a_1]) = \theta([a_2] \iff f(a_1) = f(a_2) \iff [a_1] = [a_2]$. Hence, $\theta$ is injective. $\square$.
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