# Injective and Surjective Maps

Proposition 1.Let be a morphism of closed sets, with corresponding .

- is injective if and only if is dense.
- is surjective if and only if is an embedding of V as a closed subspace of W.

**Reminder**

A continuous map of topological spaces is called an **embedding** if *f* is injective and the topology of subspace of *X* is induced via the subspace topology from *Y*. For a non-example, the map

is injective and continuous but *not* an embedding.

**Proof**

First claim: saying is injective is equivalent to: for any regular , if then . This is equivalent to: if *g* vanishes on *f*(*V*), then it vanishes on the whole of *W*.

- If this holds since
*g*is continuous. - Conversely, if , then so there exists , i.e.
*g*vanishes on*W’*(and hence*f*(*V*)) but not*W*.

Second claim: is surjective if and only if it factors through

for some ideal and isomorphism , where is the canonical map. Since *k*[*V*] is a reduced ring, such an must be a radical ideal. So the above homomorphisms correspond to: for some closed subset . ♦

## Example

Take the map , where . Algebraically

[Image edited from GeoGebra plot.]

- The image of
*f*is which is dense in ; correspondingly is injective. - Even though
*f*is injective, its image is not closed. Correspondingly is not surjective.

We will devote the remainder of this article to two rather difficult examples. Since the reasoning is a bit involved, this is all we will cover for now.

# Case Study A

Let . We take the following set

.

### Problem: is this a closed subset? What is the ideal *I*(*V*)?

For the first question, we claim that *V* is cut out by the equations and . Indeed if *x*, *y*, *z* are complex values and , and , then either *x* = 0 (in which case *y* = *z* = 0), or we set *t* := *y*/*x* to obtain

The next question is tricky; if , then we just saw that , and are in . A bit of experimentation gives us more elements

Let be the ideal generated by these 3 elements; we will show that . Now let ; we claim that *B* is spanned by , and as a *k*-vector space.

- Since and , we can replace any monomial modulo by where
*j*= 0, 1 and*k*= 0, 1. - Finally since we are done.

The morphism where corresponds to the homomorphism

Since , we obtain the composition

which maps , and into subspaces , and of . These spaces are linearly independent over *k* so the composed map is injective. Hence, the first map is injective and we see that .

**Questions to Ponder**

- Is a prime ideal of
*k*[*X*,*Y*,*Z*]? Equivalently, is*k*[*V*] an integral domain? - Can be generated by two or less elements?

# Case Study B

Let us take the subring of *k*[*S*, *T*]. Consider the *k*-algebra homomorphism:

### Problem: describe the kernel of this map.

**Step 1**. Note that the map is an injection.

Indeed, the map corresponds to and the homomorphism is an isomorphism. Now has basis given by for all . Let us compute a small set of monomial representatives such that

as vector spaces.

- Degree 1 : we have . For these we take

.

- Degree 2 : we have . But then so we leave out this term and take

.

- Degree 3 : we have . But and so there are no new terms added.

Thus for we have shown

The occurring monomials can be plotted as follows:

Note that the point (2, 2) is conspicuously absent.

**Step 2.** We write down as many relations as we can in the ideal. Some clear ones are

Do these generate the whole ideal? Let be the ideal generated by these four elements.

Strategy. We will show that is injective.

For that, we pick any .

- Replacing with and with modulo , we obtain
*f*in where the sum is over , . - Replacing with modulo , the sum is now over .
- Finally replacing with , we have where

Now suppose , then so . By looking at the monomials occuring, this can only happen for *g* = 0. This proves:

**Questions to Ponder**

- Is a prime ideal of ?
- Can be generated by three elements or less?

What do you mean by degree in the second example?

Ya I was a little vague in that part. We have the ring there. Think of it as where etc. So by degree 1, we mean degree-1 monomials in . Degree 2 just means degree-2 monomials, i.e. .

In the proof of the proposition, in the second bullet, there is a name clash between function f and element f \in I(W’) – I(W).

Thanks. Changed it to . I’ve been using too many ‘s.