Proof of the First Isomorphism Theorem

Question

Here's what I'm trying to prove;

Let $T: V \to W$ be a vector space homomorphism. Let $N = \ker(T)$. Then, the quotient vector space $V/N$ is isomorphic to $T(V)$.

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Proof Attempt:

Define the following map:

$$\forall x \in V: f(x+N) = T(x)$$

I claim that this is bijective and linear. I will prove linearity first.

1. Proof of additivity

Let $u,v \in V/N$. Then, $u = x +N$ and $v = y + N$ for some $x,y \in V$. So:

$$f(u+v) = f((x+y)+N) = T(x+y) = T(x) + T(y) = f(u) + f(v)$$

That proves additivity.

2. Proof of homogeneity

Let $\alpha \in \mathbb{F}$ and $u \in V/N$. Then, $u = x+N$ for some $x \in V$. So:

$$f(\alpha u) = f(\alpha x + N) = T(\alpha x) = \alpha T(x) = \alpha f(u)$$

That proves homogeneity. Hence, $f$ is linear.

Now, we will prove bijectivity.

1. Proof of Injectivity

Let $u,v \in V/N$ such that:

$$f(u) = f(v)$$

Now, $u = x + N$ and $v = y + N$ for some $x,y \in V$. So, the equality above implies that:

$$T(x) = T(y)$$

$$\implies T(x-y) = 0$$

$$\implies x-y \in N$$

$$\iff x + N = y+N$$

That just means that $u = v$. Hence, $f$ is injective.

2. Proof of Surjectivity

Let $w \in T(V)$. Then, there is an $x \in V$ such that $T(x) = w$. Define $u = x + N$. Clearly:

$$f(u) = f(x+N) = T(x) = w$$

So, a generic element from the codomain has a preimage under $f$. This means that $f$ is surjective.

Since $f$ is linear and bijective, it follows that it is an isomorphism between $V/N$ and $T(V)$. Hence, $V/N$ and $T(V)$ are isomorphic. That proves the desired result.

Does the proof above work? If it doesn't, why? What can I do to fix it?

Answer 1

Your proof is almost correct. But you forgot one very important step. When we define a function on equivalence classes which does something with the representatives we have to prove it is well defined, i.e does not depend on the choice of the representatives.

So in your example we have to show that if $x+N=y+N$ then $f(x+N)=f(y+N)$, otherwise it will not be a well defined function. This is easy though. Suppose $x+N=y+N$. This means $x-y\in N$, and by the definition of $N$ we get $T(x-y)=0$. Since $T$ is linear this implies $T(x)=T(y)$, so indeed $f(x+N)=f(y+N)$.