[proofplan] We first verify that $\ker\alpha$ and $\mathrm{im}\,\alpha$ are subspaces by checking closure under linear combinations. We then define the induced map $\vartheta: U/\ker\alpha \to \mathrm{im}\,\alpha$ by $\vartheta(u + \ker\alpha) = \alpha(u)$ and verify it is well-defined, linear, injective, and surjective in sequence. Well-definedness follows from the definition of the kernel, injectivity from the triviality of $\ker\vartheta$, and surjectivity from the definition of the image. The result then follows from [Isomorphism iff Linear Bijection](/theorems/378). [/proofplan] [step:Show $\ker\alpha$ is a subspace of $U$] [claim:Kernel Is Subspace] $\ker\alpha$ is a subspace of $U$. [/claim] [proof] The zero vector satisfies $\alpha(\mathbf{0}) = \mathbf{0}$, so $\mathbf{0} \in \ker\alpha$ and $\ker\alpha$ is non-empty. If $u_1, u_2 \in \ker\alpha$ and $\lambda, \mu \in \mathbb{F}$, then \begin{align*} \alpha(\lambdau_1 + \muu_2) = \lambda\alpha(u_1) + \mu\alpha(u_2) = \lambda\mathbf{0} + \mu\mathbf{0} = \mathbf{0}, \end{align*} so $\lambdau_1 + \muu_2 \in \ker\alpha$. [/proof] [/step] [step:Show $\mathrm{im}\,\alpha$ is a subspace of $V$] [claim:Image Is Subspace] $\mathrm{im}\,\alpha$ is a subspace of $V$. [/claim] [proof] Since $\alpha(\mathbf{0}) = \mathbf{0}$, we have $\mathbf{0} \in \mathrm{im}\,\alpha$. If $v_1 = \alpha(u_1)$ and $v_2 = \alpha(u_2)$ are in $\mathrm{im}\,\alpha$ and $\lambda, \mu \in \mathbb{F}$, then \begin{align*} \lambdav_1 + \muv_2 = \lambda\alpha(u_1) + \mu\alpha(u_2) = \alpha(\lambdau_1 + \muu_2) \in \mathrm{im}\,\alpha. \end{align*} [/proof] [/step] [step:Verify that $\vartheta(u + \ker\alpha) = \alpha(u)$ is well-defined] [claim:Theta Well Defined] The map $\vartheta: U/\ker\alpha \to \mathrm{im}\,\alpha$ defined by $\vartheta(u + \ker\alpha) = \alpha(u)$ is well-defined. [/claim] [proof] Suppose $u + \ker\alpha = u' + \ker\alpha$, so that $u - u' \in \ker\alpha$. Then \begin{align*} \alpha(u) - \alpha(u') = \alpha(u - u') = \mathbf{0}, \end{align*} giving $\alpha(u) = \alpha(u')$. Hence the output of $\vartheta$ does not depend on the representative of the coset. [/proof] [guided] The map $\vartheta$ is defined on cosets $u + \ker\alpha$, but the formula $\vartheta(u + \ker\alpha) = \alpha(u)$ uses a representative $u$. For this to define a function on $U/\ker\alpha$, different representatives of the same coset must give the same output. If $u + \ker\alpha = u' + \ker\alpha$, then $u - u' \in \ker\alpha$, meaning $\alpha(u - u') = \mathbf{0}$. By linearity of $\alpha$: \begin{align*} \alpha(u) - \alpha(u') = \alpha(u - u') = \mathbf{0}. \end{align*} Therefore $\alpha(u) = \alpha(u')$, confirming well-definedness. The kernel of $\alpha$ is precisely the equivalence relation that $\vartheta$ must respect -- this is why the quotient is taken by $\ker\alpha$ and not by some other subspace. [/guided] [/step] [step:Prove $\vartheta$ is $\mathbb{F}$-linear] For $u_1, u_2 \in U$ and $\lambda, \mu \in \mathbb{F}$: \begin{align*} \vartheta(\lambda(u_1 + \ker\alpha) + \mu(u_2 + \ker\alpha)) &= \vartheta((\lambdau_1 + \muu_2) + \ker\alpha) \\ &= \alpha(\lambdau_1 + \muu_2) = \lambda\alpha(u_1) + \mu\alpha(u_2) \\ &= \lambda\,\vartheta(u_1 + \ker\alpha) + \mu\,\vartheta(u_2 + \ker\alpha). \end{align*} [/step] [step:Prove $\vartheta$ is injective via triviality of its kernel] Suppose $\vartheta(u + \ker\alpha) = \mathbf{0}$. Then $\alpha(u) = \mathbf{0}$, so $u \in \ker\alpha$, whence $u + \ker\alpha = \mathbf{0} + \ker\alpha$ is the zero element of $U/\ker\alpha$. Therefore $\ker\vartheta = \{\mathbf{0} + \ker\alpha\}$, which implies $\vartheta$ is injective. [guided] To show injectivity, it suffices to show $\ker\vartheta$ is trivial (a standard criterion for injectivity of linear maps). Suppose $\vartheta(u + \ker\alpha) = \mathbf{0}$. By definition of $\vartheta$, this means $\alpha(u) = \mathbf{0}$, i.e., $u \in \ker\alpha$. But then $u + \ker\alpha = \ker\alpha$, which is the zero element of the quotient space $U/\ker\alpha$. Hence the only element of $U/\ker\alpha$ that $\vartheta$ sends to $\mathbf{0}$ is the zero coset, so $\ker\vartheta = \{\mathbf{0} + \ker\alpha\}$ and $\vartheta$ is injective. This is the payoff of quotienting by $\ker\alpha$: the ambiguity that prevented $\alpha$ from being injective has been collapsed into a single equivalence class. [/guided] [/step] [step:Prove $\vartheta$ is surjective and conclude it is an isomorphism] Every element of $\mathrm{im}\,\alpha$ has the form $\alpha(u)$ for some $u \in U$, and $\alpha(u) = \vartheta(u + \ker\alpha)$. Hence $\vartheta$ is surjective. Since $\vartheta$ is a bijective [linear map](/page/Linear%20Map), it is an isomorphism by the [Isomorphism iff Linear Bijection](/theorems/378) theorem. [/step]