A short exact sequence is defined as a sequence of algebraic objects (e.g. groups) of the form $$ 0 \longrightarrow A \overset{f}{\longrightarrow} B \overset{g}{\longrightarrow} C \longrightarrow 0$$ where $f,g$ are homomorphisms and $\text{Im}(f) = \text{Ker}(g)$. My professor said that if you have a group homomorphism $f: A\to B$, you immediately get the short exact sequence, $$ 0 \longrightarrow \text{ker}(f) \longrightarrow A \longrightarrow \text{im}(f) \longrightarrow 0. $$ Take, for instance, the homomorphism $\text{det} : \text{GL}_n (\mathbb{F}) \to \mathbb{F}^\times$, the determinant function (a multiplicative homomorphism). The kernel of the function is the set of matrices with determinant $1$: the special linear group. The image of the function is the set of non-zero elements of the field $\mathbb{F}$. So we then have the sequence, $$ 1 \longrightarrow \text{SL}_n (\mathbb{F}) \longrightarrow \text{GL}_n (\mathbb{F}) \overset{\text{det}}{\longrightarrow} \mathbb{F}^\times \longrightarrow 1$$
My question is: What exactly are the functions between $1 \overset{\phi_1}\longrightarrow \text{SL}_n (\mathbb{F})$, $\text{SL}_n (\mathbb{F}) \overset{\phi_2}\longrightarrow \text{GL}_n (\mathbb{F})$, and $\mathbb{F}^\times \overset{\phi_{3}}\longrightarrow 1$? I know that the image of each $\phi_i$ must be the kernel of the next morphism (so $\text{im} (\phi_2) = \text{ker} (\text{det})$, for example). The only thing I can think of is that perhaps $\phi_{2}$ is given by the inclusion map... Could someone clarify?
To summarize the comments given by @Randall and @user113102 (thank you!), the kernel of $f$ is obviously a subset of $A$, there is only one possibility for the functions mapping the identity to the kernel and the image to the identity: the trivial homomorphism (homomorphisms preserve identity) and the zero-map, respectively. So, for the example of the short exact sequence given by the determinant map,
$$1 \overset{\phi_1}\longrightarrow \text{SL}_n (\mathbb{F}) \overset{\phi_2}\longrightarrow \text{GL}_n (\mathbb{F}) \overset{\text{det}}\longrightarrow \mathbb{F}^\times \overset{\phi_3}\longrightarrow 1 $$
we have that $\phi_{1}: \{ 1\} \to \text{SL}_n (\mathbb{F})$ is given by $\phi_1 (1) = I$ (identity matrix), $\phi_2$ is the inclusion map, and $\phi_3 (a) = 1$ for all $a \in \mathbb{F}$, the "zero-map" (but it's multiplicative, so really the one-map).