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I assume that by a small morphism you mean a proper birational morphism $f:Y\to X$ such that $f$ does not have an exceptional divisor. If $X$ is smooth in codimension $1$ as in your last sentence, then this is equivalent to that $f$ is an isomorphism in codimension $1$ on both $Y$ and $X$. A small morphism is a small resolution if $Y$ is smooth.

There is a nice criterion to check that a variety does not admit a small resolution:

Let $f:Y\to X$ be a small morphism. If $Y$ is quasi-projective and $X$ is normal then $X$ is not $\mathbb Q$-factorial (being $\mathbb Q$-factorial means that every Weil divisor is $\mathbb Q$-Cartier, that is, it has a non-zero integer multiple which is Cartier). The proof is very simple: Let $H$ be a Cartier divisor on $Y$ that is not trivial on a curve that gets contracted by $f$, for example an ample divisor on $Y$ will do. Now if $X$ were $\mathbb Q$-factorial, then $m(f_*H)$ is a Cartier divisor for some $m\neq 0$. Then by the condition the divisors $mH$ and $f^*(m(f_*H))$ agree. However, the former was chosen to be non-trivial on a curve that is contracted while the latter must be trivial on every such curve as it is a pull-back. Q.E.D.

There is also something one can say for the reverse direction:

If $X$ has klt singularities, then it is possible to construct a morphism $f:Y\to X$ such that $f$ is small and $Y$ is $\mathbb Q$-factorial. I don't know a very easy proof of this. The essence is to take a resolution and then use a well-chosen directed mmp (in the sense of BCHM) to contract all the exceptional divisors.

It seems that the existence of a non-trivial small morphism $f:Y\to X$ where $X$ has klt singularities and $Y$ is quasi-projective is equivalent to $X$ not being $\mathbb Q$-factorial. Whether or not $X$ admits a small resolution seems to be decided whether or not the "$\mathbb Q$-factorial model" obtained by the above mentioned directed mmp is smooth or not. It is possible that it just leads to something $\mathbb Q$-factorial which is still singular and that $X$ does not admit a small resolution after all.

EDIT: Removed previous statement about the reverse direction as that was not true as stated. At this time I am not sure how to fix that statement. I replaced it with a different one that I know how to prove, but the margin is not wide enough to include a proof, so it is just stated without proof. Sorry.

I assume that by a small morphism you mean a proper birational morphism $f:Y\to X$ such that $f$ does not have an exceptional divisor. If $X$ is smooth in codimension $1$ as in your last sentence, then this is equivalent to that $f$ is an isomorphism in codimension $1$ on both $Y$ and $X$. A small morphism is a small resolution if $Y$ is smooth.

There is a nice criterion to check that a variety does not admit a small resolution:

Let $f:Y\to X$ be a small morphism. If $Y$ is quasi-projective and $X$ is normal then $X$ is not $\mathbb Q$-factorial (being $\mathbb Q$-factorial means that every Weil divisor is $\mathbb Q$-Cartier, that is, it has a non-zero integer multiple which is Cartier). The proof is very simple: Let $H$ be a Cartier divisor on $Y$ that is not trivial on a curve that gets contracted by $f$, for example an ample divisor on $Y$ will do. Now if $X$ were $\mathbb Q$-factorial, then $m(f_*H)$ is a Cartier divisor for some $m\neq 0$. Then by the condition the divisors $mH$ and $f^*(m(f_*H))$ agree. However, the former was chosen to be non-trivial on a curve that is contracted while the latter must be trivial on every such curve as it is a pull-back. Q.E.D.

There is also something one can say for the reverse direction:

If $X$ has klt singularities, then it is possible to construct a morphism $f:Y\to X$ such that $f$ is small and $Y$ is $\mathbb Q$-factorial. I don't know a very easy proof of this. The essence is to take a resolution and then use a well-chosen directed mmp (in the sense of BCHM) to contract all the exceptional divisors.

This implies that the existence of a non-trivial small morphism $f:Y\to X$ where $X$ has klt singularities and $Y$ is quasi-projective is equivalent to $X$ not being $\mathbb Q$-factorial. Whether or not $X$ admits a small resolution is then decided on whether or not there exists a directed mmp so the "$\mathbb Q$-factorial model" obtained by the above method is smooth. It is possible that all choices lead to something $\mathbb Q$-factorial which is still singular and that $X$ does not admit a small resolution after all.

EDIT: Removed previous statement about the reverse direction as that was not true as stated. At this time I am not sure how to fix that statement. I replaced it with a different one that I know how to prove, but the margin is not wide enough to include a proof, so it is just stated without proof. Sorry. However, the current statement is probably pretty close to what one might hope for.

I assume that by a small morphism you mean a proper birational morphism $f:Y\to X$ such that $f$ does not have an exceptional divisor. If $X$ is smooth in codimension $1$ as in your last sentence, then this is equivalent to that $f$ is an isomorphism in codimension $1$ on both $Y$ and $X$. A small morphism is a small resolution if $Y$ is smooth.

There is a nice criterion to check that a variety does not admit a small resolution:

Let $f:Y\to X$ be a small morphism. If $Y$ is quasi-projective and $X$ is normal then $X$ is not $\mathbb Q$-factorial (being $\mathbb Q$-factorial means that every Weil divisor is $\mathbb Q$-Cartier, that is, it has a non-zero integer multiple which is Cartier). The proof is very simple: Let $H$ be a Cartier divisor on $Y$ that is not trivial on a curve that gets contracted by $f$, for example an ample divisor on $Y$ will do. Now if $X$ were $\mathbb Q$-factorial, then $m(f_*H)$ is a Cartier divisor for some $m\neq 0$. Then by the condition the divisors $mH$ and $f^*(m(f_*H))$ agree. However, the former was chosen to be non-trivial on a curve that is contracted while the latter must be trivial on every such curve as it is a pull-back. Q.E.D.

There is also something one can say for the reverse direction:

If $X$ is smooth in codimension $1$ and is not $\mathbb Q$-factorial, then let $D\subset X$ be an effective Weil divisor that is not $\mathbb Q$-Cartier. Blowing up the ideal sheaf of $D$ produces a small morphism $f:Y\to X$ such that the pre-image (not pull-back!) of $D$ is a Cartier divisor (see II.7.13 in Hartshorne). This implies that $f$ is not an isomorphism. Now one may ask again if $Y$ is $\mathbb Q$-factorial and keep going.

It seems that the existence of a non-trivial small morphism $f:Y\to X$ where $X$ is normal and $Y$ is quasi-projective is equivalent to $X$ not being $\mathbb Q$-factorial. Whether or not $X$ admits a small resolution seems to be decided whether or not successively blowing up non-$\mathbb Q$-Cartier divisors leads to something smooth or not. It is possible that it just leads to something $\mathbb Q$-factorial which is still singular and that $X$ does not admit a small resolution after all.

I assume that by a small morphism you mean a proper birational morphism $f:Y\to X$ such that $f$ does not have an exceptional divisor. If $X$ is smooth in codimension $1$ as in your last sentence, then this is equivalent to that $f$ is an isomorphism in codimension $1$ on both $Y$ and $X$. A small morphism is a small resolution if $Y$ is smooth.

There is a nice criterion to check that a variety does not admit a small resolution:

Let $f:Y\to X$ be a small morphism. If $Y$ is quasi-projective and $X$ is normal then $X$ is not $\mathbb Q$-factorial (being $\mathbb Q$-factorial means that every Weil divisor is $\mathbb Q$-Cartier, that is, it has a non-zero integer multiple which is Cartier). The proof is very simple: Let $H$ be a Cartier divisor on $Y$ that is not trivial on a curve that gets contracted by $f$, for example an ample divisor on $Y$ will do. Now if $X$ were $\mathbb Q$-factorial, then $m(f_*H)$ is a Cartier divisor for some $m\neq 0$. Then by the condition the divisors $mH$ and $f^*(m(f_*H))$ agree. However, the former was chosen to be non-trivial on a curve that is contracted while the latter must be trivial on every such curve as it is a pull-back. Q.E.D.

There is also something one can say for the reverse direction:

If $X$ has klt singularities, then it is possible to construct a morphism $f:Y\to X$ such that $f$ is small and $Y$ is $\mathbb Q$-factorial. I don't know a very easy proof of this. The essence is to take a resolution and then use a well-chosen directed mmp (in the sense of BCHM) to contract all the exceptional divisors.

It seems that the existence of a non-trivial small morphism $f:Y\to X$ where $X$ has klt singularities and $Y$ is quasi-projective is equivalent to $X$ not being $\mathbb Q$-factorial. Whether or not $X$ admits a small resolution seems to be decided whether or not the "$\mathbb Q$-factorial model" obtained by the above mentioned directed mmp is smooth or not. It is possible that it just leads to something $\mathbb Q$-factorial which is still singular and that $X$ does not admit a small resolution after all.

EDIT: Removed previous statement about the reverse direction as that was not true as stated. At this time I am not sure how to fix that statement. I replaced it with a different one that I know how to prove, but the margin is not wide enough to include a proof, so it is just stated without proof. Sorry.