How to show that this is a bijection? I am trying to prove that the Cartesian product of two finite sets is also finite.
These are the steps:
Let S and T be finite sets:
$$S×T=\{(s,t):s∈S,t∈T\}$$by definition of set union:$$S×T=\bigcup\limits_{s∈ S}^{}( \{s\}×T)$$also the mapping:$$g_s:\{s\}×T→T$$defined by :$$g_s(s,t)=t$$ must be a bijection. But Im not able to prove the bijection.
This is my try (bijection proof) :
First I must proof thats injection:
If $t_1=t_2$ then $g_s(s_1,t_1)=g_s(s_2,t_2)$
Question 1:
I dont know how to show $(s_1,t_1)=(s_2,t_2)$?
Question 2:
I know the definition of surjective function but I dont know how to apply that to this function?
 A: Your definition of "onto" doesn't really make sense. What are $s_1, s_2$? Where did they come from? I'll explain what they must be below, but before you know this they seem to come from nowhere. And the style of that statement is not what it typically means to be onto. Being onto means that your function reaches every element of the codomain $T$, but your definition is more akin to saying $g_s$ is well defined, i.e. that the same input yields the same output. This is important, but immediately true as your definition of $g_s$ was unambiguous. Anyways, let's proceed to the questions at hand.

*

*We're trying to prove that $g_s$ is injective so let $g_s(s_1, t_1) = g_s(s_2, t_2)$. First and foremost, by the very definition of $g_s$, to define $g_s(s_1, t_1)$ we need $(s_1, t_1) \in \{s_1\} \times T$. That is, $s_1 \in \{s\}$. But the only element $\{s\}$ is $s$, so we must have $s_1 = s$. The same argument applies to $s_2$, so $s_1 = s = s_2$. This is what I meant above by showing what $s_1, s_2$ must be. They must be $s$ to even try and apply $g_s$ on them. Thus, $g_s(s_1, t_1) = g_s(s, t_1) = t_1$. Similarly, $g_s(s_2, t_2) = g_s(s, t_2) = t_2$. You assumed that $g_s(s_1, t_1) = g_s(s_2, t_2)$. Hence, $t_1 = t_2$. We have already shown that $s_1 = s_2$, so $(s_1, t_1) = (s_2, t_2)$. Thus, $g_s$ is injective.


*Take a $t \in T$. We want to show that $g_s$ reaches $t$ in its image. Indeed, we see that $g_s(s, t) = t$ so every element of $t$ is realized as an output of $g_s$. Thus, $g_s$ is surjective.
A: A function $f:A \rightarrow B$ being "onto" aka surjective means that it "hits" every element in it's "target set". Formally this means that for every $b\in B$, you can find an $a\in A$ such that you "hit" $b$ with $f$ aka $f(a)=b$. In your example $T$ is the target set so you have to show that for every $t\in T$ you can find an element $a \in \{s\} \times T$ such that $g_s(a)=t$. Note that here $a$ is of the form $a=(s,t')$ for some $t'\in T$. What could $t'$ be?
